Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left(5x-5y\right)^2-\left(4x+4y\right)^2\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
b) Ta có: \(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
c) Sửa đề: \(ax^2-ay^2-7x-7y\)
Ta có: \(ax^2-ay^2-7x-7y\)
\(=a\left(x^2-y^2\right)-7\left(x+y\right)\)
\(=\left(x+y\right)\cdot a\left(x-y\right)-7\left(x+y\right)\)
\(=\left(x+y\right)\left(ax-ay-7\right)\)
d) Ta có: \(x^4-4x^2-5\)
\(=x^4-5x^2+x^2-5\)
\(=x^2\left(x^2-5\right)+\left(x^2-5\right)\)
\(=\left(x^2-5\right)\left(x^2+1\right)\)
Bài 2: Tìm x
Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
Bài 3:
Ta có: \(\left(n+3\right)^2-\left(n-1\right)^2\)
\(=\left(n+3-n+1\right)\left(n+3+n-1\right)\)
\(=4\cdot\left(2n+2\right)\)
\(=8\left(n+1\right)⋮8\)(đpcm)
