Bài 1:
a: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+\left(9xy-9y^2\right)\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)=3\left(2x+3y\right)\left(x-y\right)\)
b: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
c: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
Bài 2:
a: \(A=\left(\dfrac{x}{x^2-25}-\dfrac{x-5}{x^2+5x}\right):\dfrac{10x-25}{x^2+5x}+\dfrac{x}{5-x}\)
\(=\left(\dfrac{x}{\left(x-5\right)\cdot\left(x+5\right)}-\dfrac{x-5}{x\left(x+5\right)}\right)\cdot\dfrac{x^2+5x}{10x-25}+\dfrac{x}{5-x}\)
\(=\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{5\left(2x-5\right)}-\dfrac{x}{x-5}\)
\(=\dfrac{x^2-x^2+10x-25}{x-5}\cdot\dfrac{1}{10x-25}-\dfrac{x}{x-5}\)
\(=\dfrac{10x-25}{10x-25}\cdot\dfrac{1}{x-5}-\dfrac{x}{x-5}\)
\(=\dfrac{1}{x-5}-\dfrac{x}{x-5}=\dfrac{-x+1}{x-5}\)
b: \(x^2+4x-5=0\)
=>\(x^2+5x-x-5=0\)
=>\(\left(x+5\right)\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được: \(A=\dfrac{-1+1}{1-5}=\dfrac{0}{-4}=0\)
