Bài 1:
a) \(\dfrac{16-\left(x+3\right)^2}{x^2-2x+1}\)
\(=\dfrac{\left(4-x-3\right)\left(4+x+3\right)}{\left(x-1\right)^2}\)
\(=\dfrac{\left(1-x\right)\left(x+7\right)}{\left(1-x\right)^2}\)
\(=\dfrac{x+7}{1-x}\)
b) \(\dfrac{x^2+4x+4}{x^2+5x+6}\)
\(=\dfrac{\left(x+2\right)^2}{x^2+2x+3x+6}\)
\(=\dfrac{\left(x+2\right)^2}{x\left(x+2\right)+3\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+2}{x+3}\)
Bài 2:
a) \(\dfrac{3xy+6}{6xy+12}\)
\(=\dfrac{3\left(xy+2\right)}{6\left(xy+2\right)}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\left(Đpcm\right)\)
b) \(\dfrac{x^2-xy}{5y^2-5xy}\)
\(=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}\)
\(=\dfrac{-x\left(y-x\right)}{5y\left(y-x\right)}\)
\(=-\dfrac{x}{5y}\)
Chỗ này hình như ghi sai đề