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.

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

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

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

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

\(=\left(x^2+5x+1\right)^2\)

mình tưởng lopw 9 mới học căn

em chịu bác lớp 7 học rui

lớp 7 đã học đâu?

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

b) \(4x-25=\left(2\sqrt{x}\right)^2-5^2=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4\left(x\ge0\right)\Rightarrow x+\sqrt{x}-4\sqrt{x}-4\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\)

b) \(4x-25=\left(2\sqrt{x}-5\right)\left(2\sqrt{x}+5\right)\)

c) \(x-3\sqrt{x}-4=\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)\)

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

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

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

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)

a) \(x-4\sqrt{x-2}+2\left(x\ge2\right)\) 

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

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

\(=\left(\sqrt{x-2}\right)^2-2\cdot2\cdot\sqrt{x-2}+2^2\)

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

b) \(x+4\sqrt{x-2}+2\left(x\ge2\right)\)

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

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

\(=\left(\sqrt{x-2}\right)^2+2\cdot2\cdot\sqrt{x-2}+2^2\)

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