\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)

\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)

Lời giải:
a.

$7-3a=(\sqrt{7}-\sqrt{3a})(\sqrt{7}+\sqrt{3a})$

b. 

$14x^2-11=(\sqrt{14}x-\sqrt{11})(\sqrt{14}x+\sqrt{11})$

c.

$3x-6\sqrt{x}-6=3(x-2\sqrt{x}-2)$
$=3[(\sqrt{x}-1)^2-3]$

$=3(\sqrt{x}-1-\sqrt{3})(\sqrt{x}-1+\sqrt{3})$

d.

$x\sqrt{x}-3\sqrt{x}-2=x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2$
$=x(\sqrt{x}-2)+2\sqrt{x}(\sqrt{x}-2)+(\sqrt{x}-2)$

$=(\sqrt{x}-2)(x+2\sqrt{x}+1)$

$=(\sqrt{x}-2)(\sqrt{x}+1)^2$

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=2+\sqrt{3}+\sqrt{6}+2\sqrt{2}\)

\(=2+\sqrt{3}+\sqrt{2}\left(2+\sqrt{3}\right)=\left(2+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)

\(2+\sqrt{3}+\sqrt{6}+\sqrt{8}=\left(\sqrt{2}+1\right)\left(2+\sqrt{3}\right)\)

a, \(x-\sqrt{x}\)= \(\sqrt{x}.\left(\sqrt{x}-1\right)\)

b, 3x+6\(\sqrt{x}\)= \(\sqrt{x}.\left(3\sqrt{x}+6\right)\)

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

d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)

=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)

=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)

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

\(x-6\sqrt{x-3}+6\text{=}x-3-6\sqrt{x-3}+9\)

\(\text{=}\left(\sqrt{x-3}\right)^2-2.3.\sqrt{x-3}+\left(3\right)^2\)

\(\text{=}\left(\sqrt{x-3}-3\right)^2\)

A =  \(x-6\)\(\sqrt{x-3}\) + 6 (đkxd \(x>3\))

A = (\(x\) - 3) - 2.3.\(\sqrt{x-3}\) + 9 

A = (\(\sqrt{x-3}\))² - 2.3.\(\sqrt{x-3}\) + 3²

A = (\(\sqrt{x-3}\)- 3)²

Lời giải:
$x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+6$

$=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)$

$=(\sqrt{x}-2)(\sqrt{x}-3)$

\(x-5\sqrt{x}+6=\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\)

Đề sai vc