7.5: Tatua usawa wa busara
- Page ID
- 176653
- Tatua milinganyo ya busara
- Tumia kazi za busara
- Tatua equation ya busara kwa variable maalum
Kabla ya kuanza, fanya jaribio hili la utayari.
- Tatua:\(\dfrac{1}{6} x+\dfrac{1}{2}=\dfrac{1}{3}\)
Ikiwa umekosa tatizo hili, kagua Mfano 2.2.9. - Tatua:
- Tatua formula\(5x+2y=10\) kwa\(y\)
Ikiwa umekosa tatizo hili, kagua Mfano 2.4.10.
Baada ya kufafanua maneno 'kujieleza' na 'equation' mapema, tumezitumia katika kitabu hiki. Tuna kilichorahisisha aina nyingi za maneno na kutatuliwa aina nyingi za equations. Tuna rahisi maneno mengi ya busara hadi sasa katika sura hii. Sasa tutasuluhisha equation ya busara.
Equation ya busara ni equation ambayo ina kujieleza busara.
Lazima uhakikishe kujua tofauti kati ya maneno ya busara na usawa wa busara. Equation ina ishara sawa.
\[\text {Rational Expression }\quad \quad \text{ Rational Equation} \nonumber \]
\[\dfrac{1}{8} x+\dfrac{1}{2} \quad \quad \dfrac{1}{8} x+\dfrac{1}{2}=\dfrac{1}{4} \nonumber \]
\[\dfrac{y+6}{y^{2}-36} \quad \quad \quad \dfrac{y+6}{y^{2}-36}=y+1 \nonumber \]
\[\dfrac{1}{n-3}+\dfrac{1}{n+4} \quad \quad \quad \quad \dfrac{1}{n-3}+\dfrac{1}{n+4}=\dfrac{15}{n^{2}+n-12} \nonumber \]
Kutatua milinganyo ya busara
Tayari tumetatua equations linear ambayo ilikuwa na sehemu ndogo. Tulipata LCD ya vipande vyote katika equation na kisha kuzidisha pande zote mbili za equation na LCD na “wazi” FRACTIONS.
Tutatumia mkakati huo wa kutatua usawa wa busara. Tutazidisha pande zote mbili za equation na LCD. Kisha, tutakuwa na equation ambayo haina maneno ya busara na hivyo ni rahisi sana kwetu kutatua. Lakini kwa sababu equation ya awali inaweza kuwa na kutofautiana katika denominator, ni lazima tuwe makini kwamba hatuwezi kuishia na suluhisho ambalo lingeweza kufanya denominator sawa na sifuri.
Hivyo kabla ya kuanza kutatua equation mantiki, sisi kuchunguza ni kwanza kupata maadili ambayo kufanya yoyote denominators sifuri. Kwa njia hiyo, tunapotatua usawa wa busara tutajua kama kuna ufumbuzi wowote wa algebraic tunapaswa kuacha.
Suluhisho la algebraic kwa equation ya busara ambayo ingeweza kusababisha maneno yoyote ya busara kuwa haijulikani inaitwa suluhisho la nje kwa equation ya busara.
Suluhisho la nje kwa equation ya busara ni suluhisho la algebraic ambalo lingeweza kusababisha maneno yoyote katika equation ya awali kuwa haijulikani.
Tunaona ufumbuzi wowote wa nje\(c\), kwa kuandika\(x\neq c\) karibu na equation.
Kutatua:\[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \]
Suluhisho
Hatua ya 1. Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
Ikiwa\(x=0\), basi\(\dfrac{1}{x}\) haijulikani. Hivyo tutaweza kuandika\(x \neq 0\) karibu na equation.
\[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6}, x \neq 0 \nonumber \]
Hatua ya 2. Kupata denominator angalau ya kawaida ya denominators wote katika equation.
Kupata LCD ya\(\dfrac{1}{x}\),\(\dfrac{1}{3}\), na\(\dfrac{5}{6}\)
LCD ni\(6x\).
Hatua ya 3. Futa sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD.
Kuzidisha pande zote mbili za equation na LCD,\(6x\).
\[{\color{red}6 x} \cdot\left(\dfrac{1}{x}+\dfrac{1}{3}\right)={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \]
Tumia Mali ya Mgawanyo.
\[{\color{red}6 x} \cdot \dfrac{1}{x}+{\color{red}6 x} \cdot \dfrac{1}{3}={\color{red}6 x} \cdot\left(\dfrac{5}{6}\right) \nonumber \]
Kurahisisha - na tazama, hakuna sehemu ndogo zaidi!
\[6+2 x=5 x \nonumber \]
Hatua ya 4. Tatua equation inayosababisha.
Kurahisisha.
\[\begin{aligned} &6=3 x\\ &2=x \end{aligned} \nonumber \]
Hatua ya 5. Angalia.
Ikiwa maadili yoyote yanayopatikana katika Hatua ya 1 ni ufumbuzi wa algebraic, uondoe. Angalia ufumbuzi wowote uliobaki katika usawa wa awali.
Hatukupata 0 kama suluhisho la algebraic.
\[\dfrac{1}{x}+\dfrac{1}{3}=\dfrac{5}{6} \nonumber \]
Sisi badala\(x=2\) katika equation ya awali.
\[\begin{aligned} \frac{1}{2}+\frac{1}{3}&\overset{?}{=}\frac{5}{6} \\ \frac{3}{6}+\frac{2}{6}&\overset{?}{=}\frac{5}{6} \\ \frac{5}{6}&=\frac{5}{6} \surd \end{aligned} \nonumber \]
Suluhisho ni\(x=2\)
Kutatua:\[\dfrac{1}{y}+\dfrac{2}{3}=\dfrac{1}{5} \nonumber \]
- Jibu
-
\(y=-\dfrac{15}{7}\)
Kutatua:\[\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x} \nonumber \]
- Jibu
-
\(x=\dfrac{15}{3}\)
Hatua za njia hii zinaonyeshwa.
- Hatua ya 1. Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
- Hatua ya 2. Kupata denominator angalau ya kawaida ya denominators wote katika equation.
- Hatua ya 3. Futa sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD.
- Hatua ya 4. Tatua equation inayosababisha.
- Hatua ya 5. Angalia:
- Ikiwa maadili yoyote yanayopatikana katika Hatua ya 1 ni ufumbuzi wa algebraic, uondoe.
- Angalia ufumbuzi wowote uliobaki katika usawa wa awali.
Sisi daima kuanza kwa kutambua maadili ambayo ingeweza kusababisha denominators yoyote kuwa sifuri.
Kutatua:\[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \]
Suluhisho
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}}, y \neq 0 \nonumber \]
Kupata denominator angalau ya kawaida ya denominators wote katika equation. LCD ni\(y^2\).
Futa sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD.
\[y^{2}\left(1-\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber \]
Kusambaza.
\[y^{2} \cdot 1-y^{2}\left(\dfrac{5}{y}\right)=y^{2}\left(-\dfrac{6}{y^{2}}\right) \nonumber \]
Kuzidisha.
\[y^{2}-5 y=-6 \nonumber \]
Tatua equation inayosababisha. Kwanza kuandika equation quadratic katika fomu ya kawaida.
\[y^{2}-5 y+6=0 \nonumber \]
Sababu.
\[(y-2)(y-3)=0 \nonumber \]
Tumia mali ya Bidhaa ya Zero.
\[y-2=0 \text { or } y-3=0 \nonumber \]
Kutatua.
\[y=2 \text { or } y=3 \nonumber \]
Angalia. Hatukupata\(0\) kama suluhisho la algebraic.
Angalia\(y=2\) na\(y=3\) katika equation ya awali.
\[1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \quad \quad \quad 1-\dfrac{5}{y}=-\dfrac{6}{y^{2}} \nonumber \]
\[1-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{2^{2}} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{3^{2}} \nonumber \]
\[1-\dfrac{5}{2} \overset{?}{=}-\dfrac{6}{4} \quad \quad \quad 1-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]
\[\dfrac{2}{2}-\dfrac{5}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad \dfrac{3}{3}-\dfrac{5}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]
\[-\dfrac{3}{2} \overset{?}{=} -\dfrac{6}{4} \quad \quad \quad -\dfrac{2}{3} \overset{?}{=} -\dfrac{6}{9} \nonumber \]
\[-\dfrac{3}{2}=-\dfrac{3}{2} \surd \quad \quad \quad -\dfrac{2}{3}=-\dfrac{2}{3} \surd \nonumber \]
Suluhisho ni\(y=2,y=3\)
Kutatua:\[1-\dfrac{2}{x}=\dfrac{15}{x^{2}} \nonumber \]
- Jibu
-
\(x=-3, x=5\)
Kutatua:\[1-\dfrac{4}{y}=\dfrac{12}{y^{2}} \nonumber \]
- Jibu
-
\(y=-2, y=6\)
Katika mfano unaofuata, denominators ya mwisho ni tofauti ya mraba. Kumbuka kuzingatia kwanza kupata LCD.
Kutatua:\[\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{x^{2}-4} \nonumber \]
Suluhisho
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[\dfrac{2}{x+2}+\dfrac{4}{x-2}=\dfrac{x-1}{(x+2)(x-2)}, x \neq-2, x \neq 2 \nonumber \]
Kupata denominator angalau ya kawaida ya denominators wote katika equation. LCD ni\((x+2)(x-2)\).
Futa sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD.
\[(x+2)(x-2)\left(\dfrac{2}{x+2}+\dfrac{4}{x-2}\right)=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \]
Kusambaza.
\[(x+2)(x-2) \dfrac{2}{x+2}+(x+2)(x-2) \dfrac{4}{x-2}=(x+2)(x-2)\left(\dfrac{x-1}{x^{2}-4}\right) \nonumber \]
Ondoa mambo ya kawaida.
\[\cancel {(x+2)}(x-2) \dfrac{2}{\cancel {x+2}}+(x+2){\cancel {(x-2)}} \dfrac{4}{\cancel {x-2}}=\cancel {(x+2)(x-2)}\left(\dfrac{x-1}{\cancel {x^{2}-4}}\right) \nonumber \]
Kurahisisha.
\[2(x-2)+4(x+2)=x-1 \nonumber \]
Kusambaza.
\[2 x-4+4 x+8=x-1 \nonumber \]
Kutatua.
\[\begin{aligned} 6 x+4&=x-1\\ 5 x&=-5 \\ x&=-1 \end{aligned}\]
Angalia: Hatukupata 2 au -2 kama ufumbuzi wa algebraic.
Angalia\(x=-1\) katika equation ya awali.
\[\begin{aligned} \dfrac{2}{x+2}+\dfrac{4}{x-2} &=\dfrac{x-1}{x^{2}-4} \\ \dfrac{2}{(-1)+2}+\dfrac{4}{(-1)-2} &\overset{?}{=} \dfrac{(-1)-1}{(-1)^{2}-4} \\ \dfrac{2}{1}+\dfrac{4}{-3} &\overset{?}{=} \dfrac{-2}{-3} \\ \dfrac{6}{3}-\dfrac{4}{3} &\overset{?}{=} \dfrac{2}{3} \\ \dfrac{2}{3} &=\dfrac{2}{3} \surd \end{aligned} \nonumber \]
Suluhisho ni\(x=-1\).
Kutatua:\[\dfrac{2}{x+1}+\dfrac{1}{x-1}=\dfrac{1}{x^{2}-1} \nonumber \]
- Jibu
-
\(x=\dfrac{2}{3}\)
Kutatua:\[\dfrac{5}{y+3}+\dfrac{2}{y-3}=\dfrac{5}{y^{2}-9} \nonumber \]
- Jibu
-
\(y=2\)
Katika mfano unaofuata, denominator ya kwanza ni trinomial. Kumbuka kuzingatia kwanza kupata LCD.
Kutatua:\[\dfrac{m+11}{m^{2}-5 m+4}=\dfrac{5}{m-4}-\dfrac{3}{m-1} \nonumber \]
Suluhisho
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri. Tumia fomu ya sababu ya denominator ya quadratic.
\[\dfrac{m+11}{(m-4)(m-1)}=\dfrac{5}{m-4}-\dfrac{3}{m-1}, m \neq 4, m \neq 1 \nonumber \]
Kupata denominator angalau ya kawaida ya denominators wote katika equation. LCD ni\((m-4)(m-1)\)
Futa sehemu ndogo kwa kuzidisha pande zote mbili za equation na LCD.
\[(m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1)\left(\dfrac{5}{m-4}-\dfrac{3}{m-1}\right) \nonumber \]
Kusambaza.
\[(m-4)(m-1)\left(\dfrac{m+11}{(m-4)(m-1)}\right)=(m-4)(m-1) \dfrac{5}{m-4}-(m-4)(m-1) \dfrac{3}{m-1} \nonumber \]
Ondoa mambo ya kawaida.
\[\cancel {(m-4)(m-1)}\left(\dfrac{m+11}{\cancel {(m-4)(m-1)}}\right)=\cancel {(m-4)}(m-1) \dfrac{5}{\cancel{m-4}}-(m-4)\cancel {(m-1)} \dfrac{3}{\cancel {m-1}} \nonumber \]
Kurahisisha.
\[m+11=5(m-1)-3(m-4) \nonumber \]
Tatua equation inayosababisha.
\[\begin{aligned} m+11&=5 m-5-3 m+12 \\ 4&=m \end{aligned} \nonumber \]
Angalia. Suluhisho pekee la algebraic lilikuwa 4, lakini tulisema kuwa 4 ingefanya denominator sawa na sifuri. Suluhisho la algebraic ni suluhisho la nje.
Hakuna ufumbuzi wa equation hii.
Kutatua:\[\dfrac{x+13}{x^{2}-7 x+10}=\dfrac{6}{x-5}-\dfrac{4}{x-2} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Kutatua:\[\dfrac{y-6}{y^{2}+3 y-4}=\dfrac{2}{y+4}+\dfrac{7}{y-1} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Equation sisi kutatuliwa katika mfano uliopita alikuwa moja tu algebraic ufumbuzi, lakini ilikuwa suluhisho extraneous. Hiyo kushoto yetu na hakuna ufumbuzi wa equation. Katika mfano unaofuata tunapata ufumbuzi wa algebraic mbili. Hapa moja au wote wawili wanaweza kuwa ufumbuzi wa nje.
Kutatua:\[\dfrac{y}{y+6}=\dfrac{72}{y^{2}-36}+4 \nonumber \]
Suluhisho
Factor denominators wote, hivyo tunaweza kutambua thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[\dfrac{y}{y+6}=\dfrac{72}{(y-6)(y+6)}+4, y \neq 6, y \neq-6 \nonumber \]
Kupata denominator angalau kawaida. LCD ni\((y-6)(y+6)\)
Futa sehemu ndogo.
\[(y-6)(y+6)\left(\dfrac{y}{y+6}\right)=(y-6)(y+6)\left(\dfrac{72}{(y-6)(y+6)}+4\right) \nonumber \]
Kurahisisha.
\[(y-6) \cdot y=72+(y-6)(y+6) \cdot 4 \nonumber \]
Kurahisisha.
\[y(y-6)=72+4\left(y^{2}-36\right) \nonumber \]
Tatua equation inayosababisha.
\[\begin{aligned} y^{2}-6 y&=72+4 y^{2}-144\\ 0&=3 y^{2}+6 y-72 \\ 0&=3\left(y^{2}+2 y-24\right) \\ 0&=3(y+6)(y-4) \\ y&=-6, y=4 \end{aligned} \nonumber \]
Angalia.
\(y=-6\)ni suluhisho la nje. Angalia\(y=4\) katika equation ya awali.
\[\begin{aligned} \dfrac{y}{y+6} &=\dfrac{72}{y^{2}-36}+4 \\ \dfrac{4}{4+6} &\overset{?}{=}\dfrac{72}{4^{2}-36}+4 \\ \dfrac{4}{10} &\overset{?}{=} \dfrac{72}{-20}+4 \\ \dfrac{4}{10} &\overset{?}{=} -\dfrac{36}{10}+\dfrac{40}{10} \\ \dfrac{4}{10} &=\dfrac{4}{10} \surd \end{aligned} \nonumber \]
Suluhisho ni\(y=4\).
Kutatua:\[\dfrac{x}{x+4}=\dfrac{32}{x^{2}-16}+5 \nonumber \]
- Jibu
-
\(x=3\)
Kutatua:\[\dfrac{y}{y+8}=\dfrac{128}{y^{2}-64}+9 \nonumber \]
- Jibu
-
\(y=7\)
Katika hali nyingine, ufumbuzi wote wa algebraic ni wa nje.
Kutatua:\[\dfrac{x}{2 x-2}-\dfrac{2}{3 x+3}=\dfrac{5 x^{2}-2 x+9}{12 x^{2}-12} \nonumber \]
Suluhisho
Tutaanza kwa kuzingatia madhehebu yote, ili iwe rahisi kutambua ufumbuzi wa nje na LCD.
\[\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)} \nonumber \]
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}=\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}, x \neq 1, x \neq-1 \nonumber \]
Kupata denominator angalau kawaida. LCD ni\(12(x-1)(x+1)\).
Futa sehemu ndogo.
\[12(x-1)(x+1)\left(\dfrac{x}{2(x-1)}-\dfrac{2}{3(x+1)}\right)=12(x-1)(x+1)\left(\dfrac{5 x^{2}-2 x+9}{12(x-1)(x+1)}\right) \nonumber \]
Kurahisisha.
\[6(x+1) \cdot x-4(x-1) \cdot 2=5 x^{2}-2 x+9 \nonumber \]
Kurahisisha.
\[6 x(x+1)-4 \cdot 2(x-1)=5 x^{2}-2 x+9 \nonumber \]
Tatua equation inayosababisha.
\[\begin{aligned} 6 x^{2}+6 x-8 x+8&=5 x^{2}-2 x+9\\ x^{2}-1&=0 \\ (x-1)(x+1)&=0 \\ x&=1 \text { or } x=-1 \end{aligned} \nonumber \]
Angalia.
\(x=1\)na\(x=-1\) ni ufumbuzi wa nje.
Equation haina ufumbuzi.
Kutatua:\[\dfrac{y}{5 y-10}-\dfrac{5}{3 y+6}=\dfrac{2 y^{2}-19 y+54}{15 y^{2}-60} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Kutatua:\[\dfrac{z}{2 z+8}-\dfrac{3}{4 z-8}=\dfrac{3 z^{2}-16 z-16}{8 z^{2}+2 z-64} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Kutatua:\[\dfrac{4}{3 x^{2}-10 x+3}+\dfrac{3}{3 x^{2}+2 x-1}=\dfrac{2}{x^{2}-2 x-3} \nonumber \]
Suluhisho
Factor denominators wote, hivyo tunaweza kutambua thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}=\dfrac{2}{(x-3)(x+1)}, x \neq-1, x \neq \dfrac{1}{3}, x \neq 3\nonumber \]
Kupata denominator angalau kawaida. LCD ni\((3 x-1)(x+1)(x-3)\).
Futa sehemu ndogo.
\[(3 x-1)(x+1)(x-3)\left(\dfrac{4}{(3 x-1)(x-3)}+\dfrac{3}{(3 x-1)(x+1)}\right)=(3 x-1)(x+1)(x-3)\left(\dfrac{2}{(x-3)(x+1)}\right) \nonumber \]
Kurahisisha.
\[4(x+1)+3(x-3)=2(3 x-1) \nonumber \]
Kusambaza.
\[4 x+4+3 x-9=6 x-2 \nonumber \]
Kurahisisha.
\[7 x-5=6 x-2 \nonumber \]
\[x=3 \nonumber \]
Suluhisho pekee la algebraic lilikuwa\(x=3\)lakini sisi alisema kwamba\(x=3\) bila kufanya denominator sawa na sifuri. Suluhisho la algebraic ni suluhisho la nje.
Hakuna ufumbuzi wa equation hii.
Kutatua:\[\dfrac{15}{x^{2}+x-6}-\dfrac{3}{x-2}=\dfrac{2}{x+3} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Kutatua:\[\dfrac{5}{x^{2}+2 x-3}-\dfrac{3}{x^{2}+x-2}=\dfrac{1}{x^{2}+5 x+6} \nonumber \]
- Jibu
-
Hakuna suluhisho.
Tumia Kazi za Mantiki
Kufanya kazi na kazi ambazo hufafanuliwa na maneno ya busara mara nyingi husababisha usawa wa busara. Tena, tunatumia mbinu sawa za kutatua.
Kwa kazi ya busara,\(f(x)=\dfrac{2 x-6}{x^{2}-8 x+15}\):
- Pata uwanja wa kazi
- Kutatua\(f(x)=1\)
- Pata pointi kwenye grafu kwenye thamani hii ya kazi.
Suluhisho
- Kikoa cha kazi ya busara ni namba zote halisi isipokuwa wale ambao hufanya usemi wa busara usiojulikana. Ili kuwapata, tutaweka denominator sawa na sifuri na kutatua.
\[\begin{aligned} x^{2}-8 x+15&=0 \\ (x-3)(x-5)&=0 \quad \text{Factor the trinomial.}\\ x-3&=0 \quad \text {Use the Zero Product Property.}\\ x-5&=0 \quad \text {Use the Zero Product Property.}\\ x=3 &\; x=5 \text{ Solve.} \end{aligned} \nonumber \]
Kikoa ni namba zote halisi isipokuwa\(x \neq 3, x \neq 5\)
- \[f(x)=1 \nonumber \]
Mbadala katika kujieleza kwa busara.
\[\dfrac{2 x-6}{x^{2}-8 x+15}=1 \nonumber \]
Sababu ya denominator.
\[\dfrac{2 x-6}{(x-3)(x-5)}=1 \nonumber \]
Kuzidisha pande zote mbili na LCD,\((x-3)(x-5)\)
\[(x-3)(x-5)\left(\dfrac{2 x-6}{(x-3)(x-5)}\right)=(x-3)(x-5)(1) \nonumber \]
Kurahisisha.
\[2 x-6=x^{2}-8 x+15 \nonumber \]
Kutatua.
\[0=x^{2}-10 x+21 \nonumber \]
Sababu.
\[0=(x-7)(x-3) \nonumber \]
Tumia mali ya Bidhaa ya Zero.
\[x-7=0 \quad x-3=0 \nonumber \]
Kutatua.
\[x=7 \quad x=3 \nonumber \]
- Thamani ya kazi ni 1 wakati\(x=7, x=3\)Hivyo pointi kwenye grafu ya kazi hii wakati\(f(x)=1\)itakuwa\((7,1),(3,1)\).
Kwa kazi ya busara,\(f(x)=\dfrac{8-x}{x^{2}-7 x+12}\)
- Pata uwanja wa kazi.
- Kutatua\(f(x)=3\).
- Pata pointi kwenye grafu kwenye thamani hii ya kazi.
- Jibu
-
- Kikoa ni namba zote halisi isipokuwa\(x \neq 3\) na\(x \neq 4\)
- \(x=2, x=\dfrac{14}{3}\)
- \((2,3),\left(\dfrac{14}{3}, 3\right)\)
Kwa kazi ya busara,\(f(x)=\dfrac{x-1}{x^{2}-6 x+5}\)
- Kutatua\(f(x)=4\).
- Pata pointi kwenye grafu kwenye thamani hii ya kazi.
- Jibu
-
- Kikoa ni namba zote halisi isipokuwa\(x \neq 1\) na\(x \neq 5\)
- \(x=\dfrac{21}{4}\)
- \(\left(\dfrac{21}{4}, 4\right)\)
Tatua Equation ya Mantiki kwa Variable Maalum
Tulipotatua equations linear, tulijifunza jinsi ya kutatua formula kwa variable maalum. Fomula nyingi zinazotumiwa katika biashara, sayansi, uchumi, na nyanja nyingine hutumia milinganyo ya busara kuiga mfano wa uhusiano kati ya vigezo viwili au zaidi. Sasa tutaona jinsi ya kutatua equation ya busara kwa kutofautiana maalum.
Tulipoanzisha fomu ya mteremko wa mteremko kutoka kwa formula yetu ya mteremko, tulifuta sehemu ndogo kwa kuzidisha na LCD.
\[\begin{aligned} m &=\frac{y-y_{1}}{x-x_{1}} \\ m\left(x-x_{1}\right) &=\left(\frac{y-y_{1}}{x-x_{1}}\right)\left(x-x_{1}\right) \quad \text{Multiply both sides of the equation by } x-x_1.\\ m\left(x-x_{1}\right) &=y-y_{1} \quad \text {Simplify.}\\ y-y_{1} &=m\left(x-x_{1}\right) \quad \text {Rewrite the equation with the y terms on the left.} \end{aligned} \nonumber \]
Katika mfano unaofuata, tutatumia mbinu sawa na formula ya mteremko ambao tulikuwa tukipata fomu ya mteremko wa equation ya mstari kupitia hatua\((2,3)\). Tutaongeza hatua moja zaidi ya kutatua\(y\).
Kutatua:\(m=\dfrac{y-2}{x-3}\) kwa\(y\).
Suluhisho
\[m=\dfrac{y-2}{x-3} \nonumber \]
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[m=\dfrac{y-2}{x-3}, x \neq 3 \nonumber \]
Wazi FRACTIONS kwa kuzidisha pande zote mbili za equation na LCD,\(x-3\).
\[(x-3) m=(x-3)\left(\dfrac{y-2}{x-3}\right) \nonumber \]
Kurahisisha.
\[x m-3 m=y-2 \nonumber \]
Sulua neno na\(y\).
\[x m-3 m+2=y \nonumber \]
Kutatua:\(m=\dfrac{y-5}{x-4}\) kwa\(y\).
- Jibu
-
\(y=m x-4 m+5\)
Kutatua:\(m=\dfrac{y-1}{x+5}\) kwa\(y\).
- Jibu
-
\(y=m x+5 m+1\)
Kumbuka kuzidisha pande zote mbili na LCD katika mfano unaofuata.
Kutatua:\(\dfrac{1}{c}+\dfrac{1}{m}=1\) kwa\(c\)
Suluhisho
\[\dfrac{1}{c}+\dfrac{1}{m}=1 \text { for } c \nonumber \]
Kumbuka thamani yoyote ya variable ambayo kufanya yoyote denominator sifuri.
\[\dfrac{1}{c}+\dfrac{1}{m}=1, c \neq 0, m \neq 0 \nonumber \]
Futa sehemu ndogo kwa kuzidisha pande zote mbili za equations na LCD,\(cm\).
\[cm\left(\dfrac{1}{c}+\dfrac{1}{m}\right)=cm(1) \nonumber \]
Kusambaza.
\[cm\left(\frac{1}{c}\right)+cm \frac{1}{m}=cm(1) \nonumber \]
Kurahisisha.
\[m+c=cm \nonumber \]
Kukusanya masharti\(c\) na haki.
\[m=cm-c \nonumber \]
Factor kujieleza juu ya haki.
\[m=c(m-1) \nonumber \]
Ili kutenganisha\(c\), kugawanya pande zote mbili na\(m-1\).
\[\dfrac{m}{m-1}=\dfrac{c(m-1)}{m-1} \nonumber \]
Kurahisisha kwa kuondoa mambo ya kawaida.
\[\dfrac{m}{m-1}=c \nonumber \]
Kumbuka kwamba hata kama sisi kutengwa\(c=0\) na\(m=0\) kutoka equation awali, ni lazima pia sasa hali hiyo\(m \neq 1\).
Kutatua:\(\dfrac{1}{a}+\dfrac{1}{b}=c\) kwa\(a\).
- Jibu
-
\(a=\dfrac{b}{c b-1}\)
Kutatua:\(\dfrac{2}{x}+\dfrac{1}{3}=\dfrac{1}{y}\) kwa\(y\)
- Jibu
-
\(y=\dfrac{3 x}{x+6}\)
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