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Question-32379




Question Number 32379 by mondodotto@gmail.com last updated on 24/Mar/18
Answered by MJS last updated on 24/Mar/18
(a−b)^5 =a^5 −5a^4 b+10a^3 b^2 −10a^2 b^3 +5ab^4 −b^5   a=(x/(4y^3 )); b=((2y)/x^2 )  a^5 =(x^5 /(1024y^(15) ))  −5a^4 b=−5(x^4 /(256y^(12) ))×((2y)/x^2 )=−((5x^2 )/(128y^(11) )) ⇒  ⇒ A=128; p=11  10a^3 b^2 =10(x^3 /(64y^9 ))×((4y^2 )/x^4 )=(5/(8xy^7 )) ⇒  ⇒ B=8
$$\left({a}−{b}\right)^{\mathrm{5}} ={a}^{\mathrm{5}} −\mathrm{5}{a}^{\mathrm{4}} {b}+\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} −\mathrm{10}{a}^{\mathrm{2}} {b}^{\mathrm{3}} +\mathrm{5}{ab}^{\mathrm{4}} −{b}^{\mathrm{5}} \\ $$$${a}=\frac{{x}}{\mathrm{4}{y}^{\mathrm{3}} };\:{b}=\frac{\mathrm{2}{y}}{{x}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{5}} =\frac{{x}^{\mathrm{5}} }{\mathrm{1024}{y}^{\mathrm{15}} } \\ $$$$−\mathrm{5}{a}^{\mathrm{4}} {b}=−\mathrm{5}\frac{{x}^{\mathrm{4}} }{\mathrm{256}{y}^{\mathrm{12}} }×\frac{\mathrm{2}{y}}{{x}^{\mathrm{2}} }=−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{128}{y}^{\mathrm{11}} }\:\Rightarrow \\ $$$$\Rightarrow\:{A}=\mathrm{128};\:{p}=\mathrm{11} \\ $$$$\mathrm{10}{a}^{\mathrm{3}} {b}^{\mathrm{2}} =\mathrm{10}\frac{{x}^{\mathrm{3}} }{\mathrm{64}{y}^{\mathrm{9}} }×\frac{\mathrm{4}{y}^{\mathrm{2}} }{{x}^{\mathrm{4}} }=\frac{\mathrm{5}}{\mathrm{8}{xy}^{\mathrm{7}} }\:\Rightarrow \\ $$$$\Rightarrow\:{B}=\mathrm{8} \\ $$
Commented by mondodotto@gmail.com last updated on 24/Mar/18
thanx
$$\mathrm{thanx} \\ $$

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