\(x\sqrt{x}+2x+\sqrt{x}+2\left(x>0\right)\)

\(=\left(x\sqrt{x}+\sqrt{x}\right)+\left(2x+2\right)\)

\(=\sqrt{x}\left(x+1\right)+2\left(x+1\right)\)

\(=\left(\sqrt{x}+2\right)\left(x+1\right)\)

a: \(x^4-2x^3+x^2-2x\)

\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)

\(=x^3\left(x-2\right)+x\left(x-2\right)\)

\(=x\left(x-2\right)\left(x^2+1\right)\)

b: \(x^4+x^3-8x-8\)

\(=\left(x^4+x^3\right)-\left(8x+8\right)\)

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3-8\right)\)

\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)

a, \(\dfrac{x^2}{4}-xy+y^2=\left(\dfrac{x}{2}\right)^2-xy+y^2=\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.y+y^2\)

\(=\left(\dfrac{x^2}{2}-y\right)^2\)

b, \(x^2+x+\dfrac{1}{4}=x^2+\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

c, \(x^2+2\sqrt{3}x+3=x^2+2\sqrt{3}x+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)

d, \(4x^2-1=\left(2x-1\right)\left(2x+1\right)\)

`x^2/4-2*x/2*y+y^2`

`=(x/2-y)^2`

`x^2+x+1/4`

`=x^2+2*x*1/2+(1/2)^2`

`=(x+1/2)^2`

`x^2+2sqrt3x+3`

`=x+2xsqrt3+sqrt3^2`

`=(x+sqrt3)^2`

`4x^2-1`

`=(2x)^2-1`

`=(2x-1)(2x+1)`

a: \(x^4+x^2+2x+6\)

\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)

\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)

Câu b đề sai nha bạn.

a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

D

a: =64x^4+16x^2y^2+y^4-16x^2y^2

=(8x^2+y^2)^2-(4xy)^2

=(8x^2+y^2-4xy)(8x^2+y^2+4xy)

b: =x^8+2x^4+1-x^4

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

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

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

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

c: =(x+1)(x^2-x+1)+2x(x+1)

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

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

d: =(x^2-1)(x^2+1)-2x(x^2-1)

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

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

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

\(x^2+2x-8\)

\(=x^2+4x-2x-8\)

\(=x^2\left(x+4\right)-2\left(x+4\right)\)

\(=\left(x^2-2\right)\left(x+4\right)\)

\(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

\(4x^2-12x+8\)

\(=4x^2-4x-8x+8\)

\(=4x\left(x-1\right)-8\left(x-1\right)\)

\(=\left(4x-8\right)\left(x-1\right)\)

\(x^2-xy-\dfrac{3}{4}y^2\)

\(=x^2-\dfrac{3}{2}xy+\dfrac{1}{2}xy-\dfrac{3}{4}y^2\)

\(=x\left(x-\dfrac{3}{2}y\right)+\dfrac{1}{2}y\left(x-\dfrac{3}{2}y\right)\)

\(=\left(x+\dfrac{1}{2}y\right)\left(x-\dfrac{3}{2}y\right)\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)