Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\)   \(\left(ĐK:x\ge1\right)\)

    \(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)

    \(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)

    \(\Leftrightarrow4.\left(x-1\right)=8x+12\)

    \(\Leftrightarrow4x-4=8x+12\)

    \(\Leftrightarrow-4x=16\)

    \(\Leftrightarrow x=-4\left(L\right)\)

Vậy \(S=\varnothing\)

\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\) 

\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\) 

\(2\sqrt{x-1}=2\sqrt{2x+3}\) 

\(\sqrt{x-1}=\sqrt{2x+3}\) 

\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\) 

\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\) 

\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\) 

\(\Rightarrow x=2\)

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

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

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

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

\(< =>x=2\)

2/ \(\Rightarrow5\sqrt{x+1}-6\sqrt{x+1}+3\sqrt{x+1}=2\sqrt{2x+3}\)

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

\(\Rightarrow2\sqrt{x+1}=2\sqrt{2x-3}\Rightarrow\sqrt{x+1}=\sqrt{2x+3}\)

\(\Rightarrow x+1=2x+3\Rightarrow x=-2\)

bài 1: 

đkxđ: x\(\ge\)0;y\(\ge\)1;z\(\ge\)2

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

\(\Leftrightarrow2\left(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}\right)=2.\frac{1}{2}\left(x+y+z\right)\)

\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)

\(\Leftrightarrow x-2\sqrt{x}+y-2\sqrt{y-1}+z-2\sqrt{z-2}=0\)

\(\Leftrightarrow x-2\sqrt{x}+1+y-1-2\sqrt{y-1}+1+z-2-2\sqrt{z-2}+1+1=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=-1\)(Vô lí)

Vậy phương trình vô nghiệm

bài 2:

đkxđ: x+1\(\ne\)0

<=>x\(\ne\)-1

\(5\sqrt{x+1}-\sqrt{36x+36}+\sqrt{9x+9}=\sqrt{8x+12}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{36.\left(x+1\right)}+\sqrt{9.\left(x+1\right)}=\sqrt{8x+12}\)

\(\Leftrightarrow5\sqrt{x+1}-6\sqrt{x+1}+3\sqrt{x+1}=\sqrt{8x+12}\)

\(\Leftrightarrow2\sqrt{x+1}=\sqrt{8x+12}\)

\(\Leftrightarrow4.\left(x+1\right)=8x+12\)

\(\Leftrightarrow4x+4=8x+12\)

\(\Leftrightarrow-4x=8\)

\(\Leftrightarrow x=-2\)(thõa mãn)

Vậy x=-2

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: \(\sqrt{x^2-6x+9}+3x=4\)

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

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

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

a) Ta có: \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\)       \(\left(ĐK:x\ge2\right)\)

        \(\Leftrightarrow\sqrt{4}.\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}.\sqrt{x-2}=20\)

        \(\Leftrightarrow2.\sqrt{x-2}+5\sqrt{x-2}-3.\sqrt{x-2}=20\)

        \(\Leftrightarrow4.\sqrt{x-2}=20\)

        \(\Leftrightarrow\sqrt{x-2}=5\)

        \(\Leftrightarrow x-2=25\)

        \(\Leftrightarrow x=27\left(TM\right)\)

Vậy \(S=\left\{27\right\}\)

a, PT <=> \(2\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9\left(x-2\right)}=20\)

\(2\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}\sqrt{x-2}=20\)

\(\left(2+5-3\right)\sqrt{x-2}=20\)

\(4\sqrt{x-2}=20\Leftrightarrow\sqrt{x-2}=5\Leftrightarrow x-2=25\Leftrightarrow x=27\)

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

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

có thể làm chi tiết hộ em chỗ pt đc 0 ạ??

`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) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

b) \(16x^2-8x+1=\left(4x\right)^2-2.4x.1+1^2=\left(4x-1\right)^2\)

c) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)

d) \(36x^2+36x+9=\left(6x\right)^2+2.6x.3+3^2=\left(6x+3\right)^2\)

e) \(x^3+27=x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)