Ребят помогите пожалуйстаP.S буду благодарен за решение. Created by monster174. algebra-ru

Ребят помогите пожалуйстаP.S буду благодарен за решение...

Artem112

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$\frac{5x^2+x-4}{x^2+x} = \frac{(x+1)(5x-4)}{x(x+1)} = \frac{5x-4}{x}$
$y= \frac{3}{x-2} +4
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y-4= \frac{3}{x-2} 
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x-2= \frac{3}{y-4} 
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x= \frac{3}{y-4} +2$
$(x- \frac{x}{1-x} ): \frac{3x^4}{x^2-2x+1} =
\frac{x(1-x)-x}{1-x} \cdot \frac{(x-1)^2}{3x^4} =
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=\frac{x-x^2-x}{1-x} \cdot \frac{(x-1)^2}{3x^4} =
\frac{x^2(x-1)^2}{3x^4(x-1)}=
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=\frac{x-1}{3x^2}=\frac{ \frac{1}{6} -1}{3\cdot (\frac{1}{6} )^2}=
\frac{ -\frac{5}{6}}{3\cdot \frac{1}{36}}=\frac{ -\frac{5}{6}}{\frac{1}{12}}=-10$
$( \frac{2}{3-x} - \frac{4x}{9-x^2} - \frac{1}{3+x} )(9+6x+x^2)=
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 \frac{2(3+x)-4x-3+x}{(3-x)(x+3)} (3+x)^2=
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 \frac{6+2x-4x-3+x}{3-x} (3+x)= \frac{-x+3}{3-x} (3+x)=3+x$
$10 \sqrt{ \frac{2}{5} } -0.5 \sqrt{160} + \sqrt{1 \frac{1}{9} } =
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\sqrt{ 40} } - \sqrt{40} + \sqrt{ \frac{10}{9} } = \sqrt{ \frac{10}{9} } = \frac{ \sqrt{10} }{3}$

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