Question

Giúp với

Answer 1

BÀi 1:

a: \(x^2y-2xy+y=y\left(x^2-2x+1\right)=y\left(x-1\right)^2\)

b: \(x^2-6x+9-9y^2\)

\(=\left(x-3\right)^2-\left(3y\right)^2\)

=(x-3-3y)(x-3+3y)

Bài 2:

a: \(x\left(x-5\right)+\left(x+1\right)\left(2-x\right)=0\)

=>\(x^2-5x+2x-x^2+2-x=0\)

=>-4x+2=0

=>-4x=-2

=>\(x=\frac24=\frac12\)

b: \(x^2\left(x-2019\right)+2019-x=0\)

=>\(\left(x-2019\right)\left(x^2-1\right)=0\)

=>(x-2019)(x-1)(x+1)=0

=>\(\left[\begin{array}{l}x-2019=0\\ x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2019\\ x=1\\ x=-1\end{array}\right.\)