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

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

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

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

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

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

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

Ta có :

\(x^4-3x^2+1\)

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

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

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

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

1/ \(x-6\sqrt{x}-8=\left(\sqrt{x}-3+\sqrt{17}\right)\left(\sqrt{x}-3-\sqrt{17}\right)\)

2/ Bài này làm gì còn phân tích được nữa.

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

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

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

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

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

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

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

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

x³ - 7x - 6

= x³ - x - 6x - 6

= x.(x² - 1) - 6.(x + 1)

= x.(x - 1).(x + 1) - 6.(x + 1)

= (x + 1).[x.(x - 1) - 6]

= (x + 1).(x² - x - 6)