\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
theo mình đề câu c là 6x2
\(x^3+6x^2+9x-xz^2=x\left(x^2-6x+9-z^2\right)\)
\(=\left(x-3-z\right)\left(x-3+z\right)\)
\(x^2-11x+30=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)=\left(x-5\right)\left(x-6\right)\)
\(4x^2-3x-1=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)
\(9x^2-7x-2=9x^2-9x+2x-2\)
\(=9x\left(x-1\right)+2\left(x-1\right)=\left(9x+2\right)\left(x-1\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x-1\right)^2-4\)
\(=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
còn lại lát mình làm tiếp
Bài 1:
a, \(x^3-2x-y^3-2y=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
b, \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
c, \(x^3+6x^2+9x-xz^2=x\left(x^2+6x+9-z^2\right)\)
\(=x\left[\left(x+3\right)^2-z^2\right]=x\left(x+3+z\right)\left(x+3-z\right)\)
Bài 2:
a, \(x^2-11x+30=x^2-5x-6x+30=x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x-6\right)\)
b, \(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
c, \(9x^2-7x-2=9x^2-9x+2x-2=9x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(9x+2\right)\)
d, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x\right)^2-2\left(x^2+x\right)+1-6\)
\(=\left(x^2+x-1\right)^2-6==\left(x^2+x-1+\sqrt{6}\right)\left(x^2+x-1-\sqrt{6}\right)\)
Bài 3:
a, \(x^2-5x+4=0\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
b, \(7x^2-6x-1=0\Leftrightarrow7x^2-7x+x-1=0\)
\(\Leftrightarrow7x\left(x-1\right)+\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{7}\end{matrix}\right.\)
