Біреуін болмаса да шығарып беріңіздерші. Created by 2233445566778899. algebra-ru

Біреуін болмаса да шығарып беріңіздерші...

NA1966 $1)\\
\frac{x-2}{x^3}=2x-x^2\\
x-2=-x^4*(x-2)\\
x^4(x-2)+(x-2)=0\\
(x-2)(x^4+1)=0\\
x=2\\\\
2)\\
\frac{x^2-x-2}{x-3}=\frac{2(x-2)}{x(x-3)}\\ 
\frac{(x-2)(x+1)}{x-3}=\frac{2(x-2)}{x(x-3)}\\
x(x-2)(x+1)=2(x-2)\\
(x-2)(x(x+1)-2)=0\\
x^2+x-2=0\\
(x+2)(x-1)=0\\
x=-2\\
x=1\\
x=2\\
3)\\
\frac{8x-4x^2}{(1-x)(x+1)}=\frac{x^3-4x}{x+1}\\
\frac{(x-2)(x+3)x}{x^2-1}=0\\
 x \neq +-1\\
 x=2\\
 x=-3\\
x=0$
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$(x^2-5x+7)^2-(x-2)(x-3)=1\\
(x-2)(x-3)(x^2-5x+7)=0\\
x=2\\
x=3\\
x^2-5x+7=0\\
D$

855 1 - сіз дурыс жаздынызы ба , тексеріндер!

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2 и 4 во вложении смотри.

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