\(\dfrac{\left(x-2\right)^2}{-1}=\dfrac{1}{2-x}\)
=>\(\dfrac{\left(x-2\right)^2}{-1}=\dfrac{-1}{\left(x-2\right)}\)
=>\(\left(x-2\right)^3=1\)
=>x-2=1
=>x=3(nhận)
\(ĐK:x\ne2\\ Có:\dfrac{\left(x-2\right)^2}{-1}=\dfrac{1}{2-x}\left(x\ne2\right)\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{-1}=\dfrac{-1}{x-2}\\ \Leftrightarrow\left(x-2\right)^3=\left(-1\right)^2=1\\ \Rightarrow x-2=\sqrt[3]{1}=1\\ \Leftrightarrow x=1+2=3\left(TM\right)\\ Vậy:S=3\)
{x-2}2 x 2-x = -1x1
{x-2}3 = 1
{x-2}3 = 13
x-2 = 1
x = 1 + 2
x = 3