5.3: Utawala wa Quotient wa Watazamaji
- Page ID
- 164585
Kwa idadi yoyote halisi\(a\) na idadi nzuri\(m\) na\(n\), wapi\(m > n\).
Utawala wa Quotient Kwa Maonyesho ni yafuatayo.
\(\dfrac{a^m }{a^n} = a^{ m−n}\)
Kumbuka: Msingi lazima iwe sawa. Matokeo yatakuwa na msingi sawa.
Wazo:
Kutoka sehemu ya mwisho,
\(x^3 = \textcolor{blue}{x \cdot x \cdot x} \qquad x^5 = \textcolor{red}{x \cdot x \cdot x \cdot x \cdot x}\)
Quotient yao
\(\dfrac{x^ 5 }{x^3} = \dfrac{\textcolor{red}{x \cdot x \cdot x \cdot x \cdot x }}{\textcolor{blue}{x \cdot x \cdot x }}= \dfrac{\textcolor{red}{\cancel{x \cdot x\cdot x \cdot x }\cdot x }}{\textcolor{blue}{\cancel{x \cdot x\cdot x }}}= \dfrac{\textcolor{red}{x \cdot x }}{1} = \textcolor{red}{x \cdot x}\).
Hivyo,\(\dfrac{x^5 }{x^3 }= x^{5−3 }= x^2\)
Kutumia utawala wa quotient wa exponents kurahisisha maneno.
- \(\dfrac{k^3 }{k^2}\)
- \(\dfrac{r^{32} }{r^{21}}\)
- \(\dfrac{\sqrt{2}^ 7 }{\sqrt{2 }^4}\)
- \(\dfrac{(−7)^9 }{(−7)^6}\)
- \(\dfrac{(x \sqrt{5})^8 }{x\sqrt{ 5}}\)
- \(\dfrac{(xy)^{18} }{(xy)^{17}}\)
Suluhisho
Ufafanuzi | Utawala wa Quotient | Msingi |
\(\dfrac{k^3 }{k^2}\) | \(k^{3−2 }= k\) | \(k\) |
\(\dfrac{r^{32} }{r^{21}}\) | \(r^{32−21 }= r^{11}\) | \(r\) |
\(\dfrac{\sqrt{2}^ 7 }{\sqrt{2 }^4}\) | \(\sqrt{2 }^{7−4 }= \sqrt{2 }^3\) | \(\sqrt{2}\) |
\(\dfrac{(−7)^9 }{(−7)^6}\) | \((−7)^{9−6 }= (−7)^3\) | \(-7\) |
\(\dfrac{(x \sqrt{5})^8 }{x\sqrt{ 5}}\) | \((x \sqrt{5})^{8−1 }= (x \sqrt{5})^7\) | \(x\sqrt{5}\) |
\(\dfrac{(xy)^{18} }{(xy)^{17}}\) | \((xy)^{18−17 }= xy\) | \(xy\) |
Kumbuka: Katika sehemu hii exponent ya nambari ilikuwa kubwa zaidi kuliko exponent ya denominator. Hiyo si mara zote kuwa kesi. Kesi ambapo exponent katika denominator ni kubwa kuliko exponent katika nambari itajadiliwa katika sehemu ya baadaye.
Tumia utawala wa quotient wa vielelezo ili kurahisisha usemi uliotolewa.
- \(\dfrac{−y ^{13} }{−y^7}\)
- \(\dfrac{(2x)^{25}}{ 2x}\)
- \(\dfrac{\sqrt{7 }^{17 }}{\sqrt{7 }^{12}}\)
- \(\dfrac{(−7)^9 }{(−7)^6}\)
- \(\dfrac{(x + y) ^{78}}{ (x + y)^{43}}\)
- \(\dfrac{\sqrt{xy }^{15 }}{\sqrt{xy }^{11}}\)