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

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

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

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

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

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

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

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

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

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

\(x-6\sqrt{x}+8\)

\(=x-2\sqrt{x}-4\sqrt{x}+8\)

\(=\sqrt{x}\left(\sqrt{x}-2\right)-4\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)\)

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

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

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

\(x+7\sqrt{x}+10=\left(\sqrt{x}+2\right)\left(\sqrt{x}+5\right)\)

( x² - x + 4 )( x² - x + 5 ) - 6

Đặt t = x² - x + 4

Đa thức đã cho trở thành

t( t + 1 ) - 6

= t² + t - 6

= t² - 2t + 3t - 6

= t( t - 2 ) + 3( t - 2 )

= ( t - 2 )( t + 3 )

= ( x² - x + 4 - 2 )( x² - x + 4 + 3 )

= ( x² - x + 2 )( x² - x + 7 )

\(x\left(x+1\right)\left(x^2+x-5\right)-6\)

\(=\left(x^2+x\right)\left(x^2+x-5\right)-6\)

\(=\left(x^2+x^2\right)^2-5\left(x^2+x\right)-6\)

\(=\left(x^2+x\right)^2+\left(x^2+x\right)-6\left(x^2+x\right)-6\)

\(=\left(x^2+x\right)\left(x^2+x+1\right)-6\left(x^2+x+1\right)\)

\(=\left(x^2+x-6\right)\left(x^2+x+1\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)