Xóa

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

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)

Vậy x=1/5 hoặc x=2

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

\(\Leftrightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(x^2-4x-y^2+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)

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

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

\(=3x^2-3y^2-2x^2+4xy-2y^2\)

\(=x^2+4xy-5y^2\)

\(=x^2+4xy+4y^2-9y^2\)

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

\(x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)

2x²y + xy² - xy

= xy(2x + y - 1)

=xy*2x+xy*y-xy*1

=xy(2x+y-1)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

a: \(4x^2-x-5=\left(4x-5\right)\left(x+1\right)\)

b: \(x^2-2x-15=\left(x-5\right)\left(x+3\right)\)

`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`