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.

$x^4+x^3-4x^2+x+1$

$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$

$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$

$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$

$=(x-1)(x^3+2x^2-2x-1)$

$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$

$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

(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)

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)`

=(x-y)²-9²

=(x-y-9)(x-y+9)

b) \(5x^3+10x^2y+5xy^2=2\left(x^3+2x^2y+xy^2\right)\)

\(=2\left(x^3+x^2y+x^2y+xy^2\right)=2\left[x^2\left(x+y\right)+xy\left(x+y\right)\right]\)

=\(2\left(x^2+xy\right)\left(x+y\right)\)

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: \(-10x^3+2x^2=0\)

\(\Leftrightarrow-2x^2\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^2\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]\)

\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)

\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)\)

\(=\left(x-z\right)\left(x-y\right)\left(-3y+3z\right)\)

\(=-3\left(x-y\right)\left(x-z\right)\left(y-z\right)\)