Question

giúp mình làm với

Answer 1

`a, 2x + 4= x^2 - 4x + 4`

`x^2 - 6x = 0`

`-> x(x-6) = 0`

`->` \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

`b, 3x - 2 = 4`

`-> 3x = 6`

`-> x = 2`

Answer 2

\(đkx\ge-2\\ \left(\sqrt{2x+4}\right)^2=\left(x-2\right)^2\\ 2x+4-x^2+4x-4=0\\ x^2+6x=0\\ x\left(x+6\right)=0\\ =>x=0\left(thoaman\right);x=-6\left(kothoaman\right)\)