Bài 1:
a) \(4x^2-y^2+4x+1=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)
b) \(-x^2+6x-9=-\left(x-3\right)^2\)
c) \(5x^3-4x^2+3xy=x\left(5x^2-4x+3y\right)\)
d) \(2x^2-2xy-7x+7y=2x\left(x-y\right)-7\left(x-y\right)=\left(x-y\right)\left(2x-7\right)\)
Bài 2:
a) \(5x\left(x-3\right)-2x+6=0\)
\(\Rightarrow5x^2-17x+6=0\)
\(\Rightarrow\left(x-3\right)\left(5x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{2}{5}\end{matrix}\right.\)
b) \(9\left(3x-2\right)=x\left(2-3x\right)\)
\(\Rightarrow-9\left(2-3x\right)=x\left(2-3x\right)\)
\(\Rightarrow x=-9\)
Bài 1:
a: Ta có: \(4x^2+4x+1-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
b: Ta có: \(-x^2+6x-9\)
\(=-\left(x^2-6x+9\right)\)
\(=-\left(x-3\right)^2\)