b: 

ĐKXĐ: x>=4

\(5\sqrt{4x-16}-\dfrac{7}{3}\cdot\sqrt{9x-36}=36-3\sqrt{x-4}\)

=>\(5\cdot2\cdot\sqrt{x-4}-\dfrac{7}{3}\cdot3\cdot\sqrt{x-4}+3\sqrt{x-4}=36\)

=>\(6\sqrt{x-4}=36\)

=>\(\sqrt{x-4}=6\)

=>x-4=36

=>x=40

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(x^2-4x-5\)

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

\(=\left(x+1\right)\left(x-5\right)\)

b, \(x^2-4xy+4y^2-36\)

\(=\left(x-2y\right)^2-36=\left(x-2y-6\right)\left(x-2y+6\right)\)

c, \(x^2-9x+20=x^2-5x-4x+20\)

\(=x\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x-4\right)\left(x-5\right)\)

d, \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)

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

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

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

a) \(x^2-4x-5=x^2-4x+4-9\)

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

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

b) \(x^2-4xy+4y^2-36\)

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

c) \(x^2-9x+20\)

= \(\left(x-5\right)\left(x-4\right)\)

d) \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)\)

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

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

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

b) \(x^2-4xy+4y^2-36=\left(x-2y\right)^2-36=\)

c) \(x^2-9x+20=x^2-4x-5x+20=x\left(x-4\right)-5\left(x-4\right)=\left(x-5\right)\left(x-4\right)\)

d) \(\left(9x^2-36\right)-\left(3x-6\right)\left(2x+5\right)=9x^2-36-\left(6x^2+15x-12x-30\right)\)

= \(9x^2-36-6x^2-15x+12x+30=3x^2-3x-6\)

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

\(ĐK:x\ge1\\ PT\Leftrightarrow12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\\ \Leftrightarrow4\sqrt{x-1}=16\\ \Leftrightarrow\sqrt{x-1}=4\\ \Leftrightarrow x-1=16\\ \Leftrightarrow x=17\left(tm\right)\)

\(< =>2\sqrt{36\left(x-1\right)}-\dfrac{1}{3}\sqrt{9\left(x-1\right)}-4\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16 \)\(< =>12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\)
\(< =>4\sqrt{x-1}=16\)
\(< =>\sqrt{x-1}=4 \)
\(< =>x-1=16\)
\(< =>x=17\)

h) 

ĐKXĐ: $x\geq -5$

PT $\Leftrightarrow \sqrt{x+5}=6$

$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)

i) ĐKXĐ: $x\geq 5$

PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)

\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)

j) 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$

$\Leftrightarrow -2\sqrt{2x}+4=0$

$\Leftrightarrow \sqrt{2x}=2$

$\Rightarrow x=2$ (thỏa mãn)

k) ĐK: $x^2\geq 5$

PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$

$\Leftrightarrow 2\sqrt{x^2-5}=4$

$\Leftrightarrow \sqrt{x^2-5}=2$

$\Rightarrow x^2-5=4$

$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)

l) ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$

$\Leftrightarrow 4\sqrt{x+1}=4$

$\Leftrightarrow \sqrt{x+1}=1$

$\Rightarrow x+1=1$

$\Rightarrow x=0$

m) 

ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$

$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}=16$

$\Leftrightarrow \sqrt{x+1}=4$

$\Rightarrow x=15$ (thỏa mãn)

`2sqrt{36x-36}-1/3sqrt{9x-9}-4sqrt{4x-4}+sqrt{x-1}=16`

`ĐK:x>=1`

`pt<=>2sqrt{36(x-1)}-1/3sqrt{9(x-1)}-4sqrt{4(x-1)}+sqrt{x-1}=16`

`<=>12sqrt{x-1}-sqrt{x-1}-8sqrt{x-1}+sqrt{x-1}=16`

`<=>4sqrt{x-1}=16`

`<=>sqrt{x-1}=4`

`<=>x-1=16`

`<=>x=17(tmđk)`

Vậy `S={17}`

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

\(=x^2\left(x+9\right)-4\left(x+9\right)\)

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

a: Ta có: \(\sqrt{x}< 3\)

nên \(0\le x< 9\)

b: Ta có: \(\sqrt{4x+16}+\sqrt{x+4}+2\sqrt{9x+36}=35\)

\(\Leftrightarrow2\sqrt{x+4}+\sqrt{x+4}+6\sqrt{x+4}=35\)

\(\Leftrightarrow\sqrt{x+4}=\dfrac{35}{9}\)

\(\Leftrightarrow x+4=\dfrac{1225}{81}\)

hay \(x=\dfrac{901}{81}\)

a) \(\sqrt{x}< 3\Rightarrow x< 9\)

b) \(\sqrt{4x+16}+\sqrt{x+4}+2\sqrt{9x+36}=35\)

\(\Rightarrow2\sqrt{x+4}+\sqrt{x+4}+6\sqrt{x+4}=35\)

\(\Rightarrow\sqrt{x+4}=\dfrac{35}{9}\)

\(\Rightarrow x+4=\dfrac{1225}{81}\)

\(\Rightarrow x=\dfrac{901}{81}\)

c) \(\sqrt{x+2\sqrt{x-1}}=3\)

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

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

\(\Rightarrow\sqrt{x^2}=3\)

\(\Rightarrow\left|x\right|=3\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)