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![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$x^4+10x^3+26x^2+10x+1$
$=(x^4+10x^3+25x^2)+x^2+10x+1$
$=(x^2+5x)^2+2(x^2+5x)+1-x^2$
$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$
$=(x^2+4x+1)(x^2+6x+1)$
b.
$x^4+x^3-4x^2+x+1$
$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$
$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$
$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$
$=(x-1)(x^3+2x^2-2x-1)$
$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$
$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x - 4)(x2 + 4x + 16) - x(x2 - 6) = 2
x3 - 64 - x3 + 6x = 2
6x = 2 + 64
6x = 66
x = 66 : 6
x = 11
x3 - 27 + 3x(x - 3)
= (x - 3)(x2 + 3x + 9) + 3x(x - 3)
= (x - 3)(x2 + 3x + 9 + 3x)
= (x - 3)(x2 + 6x + 9)
= (x - 3)(x + 3)2
5x3 - 7x2 + 10x - 14
= 5x(x2 + 2) - 7(x2 + 2)
= (x2 + 2)(5x - 7)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(8x+4x^2-12xy\)
\(=4x\left(2+x-3y\right)\)
b) Ta có: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c) Ta có: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)^2\)
d) Ta có: \(x^2-8x-9\)
\(=x^2-9x+x-9\)
\(=\left(x-9\right)\left(x+1\right)\)
a. `8x+4x^2-12xy=4x(2+x-3y)`
b) `5x^3-10x^2+5x=5x(x^2-2x+1)`
c) `x^3+x^2y-xy^2-y^3=x^2(x+y)-y^2(x+y)=(x+y)(x^2-y^2)=(x+y)^2 (x-y)`
d) `x^2-8x-9=(x^2-2.x.4+4^2)-25=(x-4)^2-5^2=(x+1)(x-9)`
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(5x^3+10x^2y+5xy^2=2\left(x^3+2x^2y+xy^2\right)\)
\(=2\left(x^3+x^2y+x^2y+xy^2\right)=2\left[x^2\left(x+y\right)+xy\left(x+y\right)\right]\)
=\(2\left(x^2+xy\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(-10x^3+2x^2=0\)
\(\Rightarrow-2x^2\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(5x\left(x-2016\right)-x+2016=0\)
\(\Rightarrow5x\left(x-2016\right)-\left(x-2016\right)=0\)
\(\Rightarrow\left(x-2016\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=\dfrac{1}{5}\end{matrix}\right.\)
a: Ta có: \(-10x^3+2x^2=0\)
\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^2\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)\)
\(=\left(x-z\right)\left(x-y\right)\left(-3y+3z\right)\)
\(=-3\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^4+5x^3+10x-4\)
\(=\left(x^4+2x^2\right)+\left(5x^3+10x\right)-\left(2x^2+4\right)\)
\(=x^2\left(x^2+2\right)+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-2\right)\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\right]\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2\right]\)
\(=\left(x^2+2\right)\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x^2+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
b) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)