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

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

`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}`

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

Sửa đề: \(2\sqrt{36x-36}-\dfrac{1}{3}\sqrt{9x-9}-4\sqrt{4x-4}+\sqrt{x-1}=16\)

\(\Leftrightarrow12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\)

=>4 căn x-1=16

=>căn x-1=4

=>x-1=16

=>x=17

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

\(\Rightarrow x+1=2-x\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

a) \(ĐKXĐ:-1\le x\le2\)

Bình phương 2 vế ta có: 

\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )

Vậy \(x=\frac{1}{2}\)

b) \(ĐKXĐ:x\ge1\)

\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)

\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)

\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )

Vậy \(x=65\)

c) \(ĐKXĐ:x\ge1\)

\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)

\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)

\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)

\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )

Vậy \(x=5\)

\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)

\(\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)

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

\(\sqrt{x-1}\left(6-3-2+1\right)=16\)

\(2\sqrt{x-1}=16\)

\(\sqrt{x-1}=8\)

\(\left(\sqrt{x-1}\right)^2=8^2\)

\(x-1=64\)

\(x=64+1=65\)

\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)ĐK x lớn hơn hoặc bằng 1

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

\(2\sqrt{x-1}=16\)

\(\sqrt{x-1}=8\)

\(x-1=64\)

\(x=65\)thỏa mãn

chết mk quên ĐKXĐ

a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)

TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)

TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)

Vậy x=0,5...

d, đk \(x\ge-1\)

=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)

\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)

a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b) Ta có: \(\sqrt{4x^2-4x+1}=3\)

\(\Leftrightarrow\left|2x-1\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)

\(\Leftrightarrow\left|x-3\right|=4-3x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)

\(=2\sqrt{3}-4\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)

Bài 5: 

\(\widehat{B}=60^0\)

\(AB=8\sqrt{3}\left(cm\right)\)

\(BC=16\sqrt{3}\left(cm\right)\)

\(2.< =>5\sqrt{x-1}-6\sqrt{x-1}-3\sqrt{x-1}=2\sqrt{2x-3}\)

\(< =>\sqrt{x-1}\left(5-6+3\right)=2\sqrt{2x-3}\)

\(< =>2\sqrt{x-1}=2\sqrt{2x-3}\)

\(< =>x-1=2x-3\)

\(< =>x=2\)