Bài 1: Thu gọn
a) Ta có: \(A=\left(x+3\right)^2-\left(x+2\right)^2\)
\(=\left[\left(x+3\right)-\left(x+2\right)\right]\cdot\left[\left(x+3\right)+\left(x+2\right)\right]\)
\(=\left(x+3-x-2\right)\left(x+3+x+2\right)\)
\(=2x+5\)
b) Ta có: \(B=\left(2x-1\right)\left(2x+1\right)-\left(2x+3\right)^2\)
\(=4x^2-1-\left(4x^2+12x+9\right)\)
\(=4x^2-1-4x^2-12x-9\)
\(=-12x-10\)
Bài 2: Tìm x
a) Ta có: \(x^2+2x+1=9\)
\(\Leftrightarrow\left(x+1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;2\right\}\)