Bài 1 :
Câu a :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
Câu b :
\(5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2\right)-20z^2\)
\(=5\left(x-y\right)^2-20z^2\)
\(=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
Câu c :
\(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
Bài 2 :
\(5x\left(x-1\right)=x-1\)
\(5x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(5x-1\right)=0\)
\(\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=1\) và\(x=\dfrac{1}{5}\)
Câu b :
\(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(x+5\right)\left(2-x\right)=0\)
\(\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(x=-5\) và \(x=2\)
Câu c :
\(x^2-2x-3=0\)
\(x^2+x-3x-3=0\)
\(x\left(x+1\right)-3\left(x+1\right)=0\)
\(\left(x+1\right)\left(x-3\right)=0\)
\(\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy \(x=-1\) và \(x=3\)
