помогите пожалуйста дам 100 б. Created by mahmudovaulduz70. algebra-ru

Ответ:вот решение с ответом на фото

помогите пожалуйста дам 100 б...

DimaPuchkov

$(\frac{x^{-3}\cdot a^4}{16} )^{-2} \cdot (\frac{4}{x^{-2}\cdot a^3 })^{-3}=(\frac{16}{x^{-3}\cdot a^4} )^{2} \cdot (\frac{x^{-2}\cdot a^3 }{4})^{3}=(\frac{16x^3}{a^4} )^{2} \cdot (\frac{ a^3 }{4x^2})^{3}=\\ \\ = (\frac{4^2x^3}{a^4} )^{2} \cdot (\frac{ a^3 }{4x^2})^{3}= \frac{4^4x^6}{a^8} \cdot \frac{ a^9 }{4^3x^6}=4^1\cdot x^0\cdot a^1=4a$

б)

$-0,2a^3b^4\cdot 5a^2b^3=-5^{-1}a^3b^4\cdot 5a^2b^3=-5^0\cdot a^5b^7=-1\cdot a^5b^7=-a^5b^7$

$\frac{81^{25}}{27^{33}} =\frac{(3^4)^{25}}{(3^3)^{33}}=\frac{3^{100}}{3^{99}}=3^1=3$

б)

$\frac{15a^2-10ab}{3ab-2b^2}= \frac{5a\cdot(3a-2b)}{b\cdot (3a-2b)} =\frac{5a}{b}$

$a=-2, \ \ b=-0,1 \\ \\ \frac{5\cdot (-2)}{-0,1}=\frac{-10}{-10^{-1}}=10^{2}=100$

$\frac{x^6+x^4}{x^4+x^2}=\frac{x^4\cdot(x^2+1) }{x^2\cdot (x^2+1)} =x^2$

б)

$\frac{a^2+3a+9}{27-a^3} =\frac{a^2+3a+9}{3^3-a^3} =\frac{a^2+3a+9}{(3-a)\cdot (9+3a+a^2)} =\frac{1}{3-a}=-\frac{1}{a-3}$

в)

$\frac{b^7-b^{10}}{b^9-b^3}=\frac{b^7\cdot (1 -b^3)}{b^3\cdot (b^6-1)} =\frac{b^4\cdot (1-b)\cdot (1+b+b^2)}{(b^3)^2-1^2}=\frac{b^4\cdot (1-b)\cdot (1+b+b^2)}{(b^3-1)\cdot (b^3+1)}=\frac{b^4\cdot (1-b)\cdot (1+b+b^2)}{(b-1)\cdot (b^2+b+1)\cdot (b^3+1)}= \\ \\ =-\frac{b^4}{b^3+1}$

$(\frac{p}{p^2-4} +\frac{2}{2-p} + \frac{1}{p+2} ):(p-2+\frac{10-p^2}{p+2} )=(\frac{p}{(p-2)\cdot (p+2)} -\frac{2}{p-2} + \frac{1}{p+2} ):(\frac{p^2+2p-2p-4+10-p^2}{p+2} )=\\ \\ =(\frac{p-2p-4+p-2}{(p-2)\cdot (p+2)} ):(\frac{6}{p+2} )=(\frac{-6}{(p-2)\cdot (p+2)} )\cdot (\frac{p+2}{6} )=-\frac{3}{p-2}=\frac{3}{2-p}$

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