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5.5: Utawala Mbaya

  • Page ID
    164587
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    Katika kifungu cha 5.3, kielelezo cha nambari katika namba kilikuwa kikubwa zaidi kuliko kielelezo cha idadi katika denominator. Katika kifungu cha 5.4, kielelezo cha nambari katika namba kilikuwa sawa na kielelezo cha idadi katika denominator. Katika kifungu cha 5.5, kielelezo cha idadi katika denominator inaweza kuwa kubwa zaidi kuliko kielelezo cha idadi katika nambari.

    Ufafanuzi: Utawala wa Maonyesho mabaya

    Kwa yoyote zero zero halisi idadi a na n yoyote integer, hasi exponent utawala ni yafuatayo

    \(a^{−n}= \dfrac{1 }{a^n} or \dfrac{1 }{a^{−n}} = a^n\)

    Ni fomu duni katika hisabati kuacha exponents hasi katika jibu. Majibu yote daima kuwa rahisi kuonyesha exponents chanya.

    Je! Hii inafanya kazi gani?

    Kumbuka:

    \[\begin{align*} 2^3 &= 2 \cdot 2 \cdot 2 = 8 \\[4pt] 2^2 &= 2 \cdot 2 = 4 \\[4pt] 2^1 &= 2 = 2 \\[4pt] 2^0 &= 1 \end{align*} \nonumber \]

    Nini kinatokea kwa exponents hasi?

    \[\begin{align*} 2^{−1 } &= \dfrac{1 }{2^1 }= \dfrac{1 }{2} \\[4pt] 2^{−2 } &= \dfrac{1}{2^2} = \dfrac{1}{ 4} \end{align*} \nonumber \]

    Kumbuka: Kutoka sehemu ya mwisho,

    \[\begin{align*} x^3 = \textcolor{blue}{x \cdot x \cdot x} \\[4pt] x^5 &= \textcolor{red}{x \cdot x \cdot x \cdot x \cdot x} \end{align*} \nonumber \]

    Quotient yao:

    \(\dfrac{x^3 }{x^5} = \dfrac{x \cdot x \cdot x }{x \cdot x \cdot x \cdot x \cdot x }= \dfrac{\textcolor{blue}{\cancel{x \cdot x\cdot x }}}{\textcolor{red}{\cancel{x \cdot x\cdot x \cdot x \cdot x }}}=\dfrac{ 1 }{\textcolor{red}{x \cdot x }}= \dfrac{1 }{x^2}\)

    Tumia utawala wa quotient ili kupata matokeo sawa.

    \(\dfrac{x^3 }{x^5} = x^{3−5 }= x^{−2}\)

    Kwa kutumia hasi exponent utawala.

    \(x^{−2 }= \dfrac{1 }{x^2}\).

    Tathmini mifano ifuatayo ili kusaidia kuelewa mchakato wa kurahisisha kutumia utawala wa quotient wa watazamaji na utawala hasi exponent.

    Kidokezo: Kuwa na subira, kuchukua muda wako na kuwa makini wakati wa kurahisisha!

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{t^5}{ t^11}\)

    Suluhisho

    \(t^{5−11 }= t^{−6 }= \dfrac{1 }{t^6}\)

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{x^{3} \cdot x^{11} }{x \cdot x^{15}}\)

    Suluhisho

    \(\dfrac{x^{ 3+11 }}{x^{1+15 }}= \dfrac{x^14 }{x^16 }= x^{14−16 }= x^{−2 }= \dfrac{1}{x^2}\)

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{2y^3 }{7y^7}\)

    Suluhisho

    \(\dfrac{2 }{7} \cdot \dfrac{y^3 }{y^7 }= \dfrac{2 }{7} \cdot y^{3−7 }= \dfrac{2 }{7 }\cdot y^{−4} = \dfrac{2 }{7 }\cdot \dfrac{1 }{y^4 }= \dfrac{2 }{7y^4}\)

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(-\dfrac{\sqrt{3}z^6}{ z^7}\)

    Suluhisho

    \(− \sqrt{3} \cdot \dfrac{z^6 }{z^7} = − \sqrt{3} \cdot z^{6−7 }= − \sqrt{3} \cdot z^{−1} = − \sqrt{3} \cdot \dfrac{1 }{z} = − \dfrac{\sqrt{3}}{z}\)

    Katika Mifano 3 na 4, sababu nje ya mara kwa mara kuona besi ya kawaida wazi.

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{1}{ a^{−9}}\)

    Suluhisho

    \(a^9\)

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{x^3 }{x^{−5}}\)

    Suluhisho

    \(x^{3−(−5) }= x^{3+5 }= x^{8}\)

    Kurahisisha kujieleza zifuatazo kwa msingi moja na exponents chanya tu.

    \(\dfrac{c^{−7 }}{c^{−3}}\)

    Suluhisho

    \(c^{(−7)−(−3) }= c^{−7+3 }= c ^{−4} = \dfrac{1 }{c^4}\)

    Katika mifano 6 na 7, utawala wa quotient wa exponents ulitumiwa kabla ya kubadilisha exponents kwa exponents chanya. Matokeo sawa ni kupatikana kwa kupanua na kubadilisha exponents kwa exponents chanya kwanza na kisha kutumia utawala quotient ya exponents.

    Kurahisisha maneno yafuatayo kwa msingi mmoja na exponents chanya tu.

    1. \(\dfrac{p^4}{ p^{13}}\)
    2. \(-\dfrac{k^2 \cdot k^3 }{k^7 \cdot k^8}\)
    3. \(\dfrac{5(x + y)^3 }{2(x + y)^{13}}\)
    4. -\(\dfrac{\sqrt{8}y^3}{ y^{−3}}\)
    5. \(\dfrac{a^{−7}}{ a^2 \cdot a^{−5}}\)
    6. \(\dfrac{x^{−7}}{ x^5}\)