Bài 1:

Tại $x=-1$ thì giá trị biểu thức là:

$x^2-4x+4=\left(x-2\right)^2=\left(-1-2\right)^2=9$

Bài 2:

$A=\left(x-1\right)^2+\left(x+1\right)\left(3-x\right)$

$=x^2-2x+1+3\left(x+1\right)-x\left(x+1\right)$

$=x^2-2x+1+3x+3-x^2-x$

$=4$. Vậy giá trị biểu thức A ko phụ thuộc vào biến

Bài 3:

a)$2x-\left(x-3\right)^2=5-x$

$\Leftrightarrow2x-\left(x-3\right)^2-5+x=0$

$\Leftrightarrow3x-5-\left(x-3\right)^2=0$

$\Leftrightarrow3x-5-x^2+6x-9=0$

$\Leftrightarrow-x^2+9x-14=0$

$\Leftrightarrow\left(7-x\right)\left(x-2\right)=0$

$\Rightarrow\left[{}\begin{matrix}7-x=0\\x-2=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.$

b)$\left(5x-2x\right)^2-\left(x+2\right)^2=0$

$\Leftrightarrow\left(5x-2x+x+2\right)\left[\left(5x-2x\right)-\left(x+2\right)\right]=0$

$\Leftrightarrow2\left(2x+1\right)\left[\left(5x-2x\right)-\left(x+2\right)\right]=0$

$\Leftrightarrow4\left(2x+1\right)\left(x-1\right)=0$

$\Rightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.$$\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.$