a) x3 - x + 3x2y + 3xy2 + y3 - y
=(x3 + 3x2y + 3xy2 + y3) - ( x + y )
=(x+y)3 - (x+y)
=(x+y)(x2+2xy+y2-1) = (x+y)(x+y-1)(x+y+1)
Bài 2a) 5x (x - 1) = x - 1
<=> 5x (x - 1) - (x - 1) = 0
<=> (x - 1)(5x - 1) = 0
[\(\begin{matrix}x-1=0\\5x-1=0\end{matrix}\)=> [\(\begin{matrix}x=1\\5x=1\end{matrix}\)=>[\(\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\)
Vậy x = 1 và x = \(\dfrac{1}{5}\)
c) 5x2+5xy-x-y
=(5x2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)
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 \(x=3\)
Bài 1b) 5x2-10xy+5y2-20z2
= 5(x2-2xy-y2) - 20z2
= 5(x-y)2 - 20z2
= 5[(x-y)2 - 4z2]
= 5(x-y-2z)(x-y-2z)