Question

Answer 1

Có cả điều kiện ạ

Answer 2

b: Ta có: \(\sqrt{x^2-2x}=2-x\)

\(\Leftrightarrow x^2-2x=x^2+4x+4\)

\(\Leftrightarrow-6x=4\)

hay \(x=-\dfrac{2}{3}\left(tm\right)\)

c: Ta có: \(\sqrt{2x+27}-6=x\)

\(\Leftrightarrow2x+27=x^2+12x+36\)

\(\Leftrightarrow x^2+10x+9=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

 

Answer 3

b: Ta có: \(\sqrt{x^2-2x}=2-x\)

\(\Leftrightarrow x^2-2x=x^2-4x+4\)

\(\Leftrightarrow2x=4\)

hay x=2