Câu 1:
a: \(2x^2+6x\)
\(=2x\cdot x+2x\cdot3\)
=2x(x+3)
b: \(x^2-4x+4-y^2\)
\(=\left(x-2\right)^2-y^2\)
=(x-2-y)(x-2+y)
Câu 2:
a: \(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\)
\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}\)
\(=x^2+3\)
b: \(\frac{2x}{x+3}-\frac{x-3}{x+3}\)
\(=\frac{2x-x+3}{x+3}=\frac{x+3}{x+3}=1\)
c: \(\frac{x}{x-2}+\frac{2-x}{x+2}+\frac{12-10x}{x^2-4}\)
\(=\frac{x}{x-2}-\frac{x-2}{x+2}+\frac{12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x+2\right)-\left(x-2\right)^2+12-10x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x-x^2+4x-4+12-10x}{\left(x-2\right)\left(x+2\right)}=\frac{-4x+8}{\left.\left(x-2\right)\left(x+2\right)\right.}\)
\(=\frac{-4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-4}{x+2}\)
Câu 3:
a: Xét tứ giác AEHF có \(\hat{AEH}=\hat{AFH}=\hat{FAE}=90^0\)
nên AEHF là hình chữ nhật
