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.

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

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

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

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

\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)

\(ĐKXĐ:x;y\ge0\)

\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

\(=\left(\sqrt{x^3}+\sqrt{x^2y}\right)-\left(\sqrt{y^3}+\sqrt{xy^2}\right)\)

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

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

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

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

mình tưởng lopw 9 mới học căn

em chịu bác lớp 7 học rui

lớp 7 đã học đâu?

a/ \(x^2+5\sqrt{x}+6=x^2+2\sqrt{x}+3\sqrt{x}+6\)

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

b/ \(x^2+4\sqrt{x}+3=x^2+\sqrt{x}+3\sqrt{x}+3\)

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

c/ k bik làm

Bạn trả lời sai rồi. \(\sqrt{x}\cdot\sqrt{x}\)chỉ được \(x\)mà thôi, ko được \(x^2\)đâu!

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

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

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

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

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

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

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

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

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

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

2

\(M=2y-3x\sqrt{y}+x^2=y-2x\sqrt{y}+x^2+y-x\sqrt{y}\\ =\left(\sqrt{y}-x\right)^2+\sqrt{y}\left(\sqrt{y}-x\right)\\ =\left(\sqrt{y}-x\right)\left(\sqrt{y}-x+\sqrt{y}\right)\\ =\left(\sqrt{y}-x\right)\left(2\sqrt{y}-x\right)\)

b

\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)

Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:

\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)

1:

a: =>2x-2căn x+3căn x-3-5=2x-4

=>căn x-8=-4

=>căn x=4

=>x=16

b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>(căn x-2)(x-căn x+4)=0

=>căn x-2=0

=>x=4