`\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12`      `ĐK: x >= 1/2`

`<=>2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12`

`<=>4\sqrt{2x-1}=12`

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

`<=>2x-1=9`

`<=>x=5` (t/m)

Vậy `S={5}`.

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

=>\(4\sqrt{2x-1}=12\)

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

=>2x-1=9

=>2x=10

=>x=5

\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

\(\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\)

\(\Leftrightarrow\left(2-6+8\right)\sqrt{2x-1}=12\)

\(\Leftrightarrow4\sqrt{2x-1}=12\)

\(\Leftrightarrow\sqrt{2x-1}=12:4\)

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

\(\Leftrightarrow2x-1=9\)

\(\Leftrightarrow2x=9+1\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(x=5\)

ai trả lời dùm em cái ak. E cảm ơn nhiều

a.\(\sqrt{x-2}=\sqrt{4-x}\)

đk: \(\left\{{}\begin{matrix}x-2\ge0\\4-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\le4\end{matrix}\right.\Leftrightarrow2\le x\le4\)

pt đã cho tương đương với

\(x-2=4-x\)

\(\Leftrightarrow2x=6\Rightarrow x=3\left(TM\right)\)

b.\(\sqrt{x^2-8x+6}=x+2\)

đk: \(x+2\ge0\Rightarrow x\ge-2\)

pt đã cho tương đương với

\(x^2-8x+6=\left(x+2\right)^2\)

\(\Leftrightarrow x^2-8x+6=x^2+4x+4\)

\(\Leftrightarrow-12x=-2\Rightarrow x=\frac{1}{6}\left(TM\right)\)

c.\(\sqrt{2x-1}+5=\sqrt{8x-4}\)

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

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

\(\Leftrightarrow\sqrt{2x-1}=5\)

đk: \(2x-1\ge0\Leftrightarrow x\ge\frac{1}{2}\)

pt tương đương: \(2x-1=25\)

\(\Leftrightarrow2x=26\Rightarrow x=13\left(TM\right)\)

d.\(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)

\(\Leftrightarrow\sqrt{16\left(1-2x\right)}-\sqrt{4.3x}=\sqrt{3x}+\sqrt{9\left(1-2x\right)}\)

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

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

đk: \(\left\{{}\begin{matrix}1-2x\ge0\\3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{1}{2}\\x\ge0\end{matrix}\right.\Leftrightarrow0\le x\le\frac{1}{2}\)

pt tương đương: \(1-2x=9.3x\)

\(\Leftrightarrow29x=1\Rightarrow x=\frac{1}{29}\left(TM\right)\)

e. \(\sqrt{x^2-9}-\sqrt{4x-12}=0\)

đk: \(\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ge0\\4x-12\ge0\end{matrix}\right.\Leftrightarrow x\ge3\)

pt đã cho tương đương với

\(\sqrt{\left(x-3\right)\left(x+3\right)}-\sqrt{4\left(x-3\right)}=0\)

\(\Leftrightarrow\sqrt{x-3}.\sqrt{x+3}-2\sqrt{x-3}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Rightarrow x=3\left(TM\right)\\\sqrt{x+3}=2\Leftrightarrow x+3=4\Rightarrow x=1\left(KTM\right)\end{matrix}\right.\)

phần a đây nhé \(a,\sqrt{4\left(2x-1\right)}-2\sqrt{9\left(2x-1\right)}+2\sqrt{16\left(2x-1\right)}=12\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\Leftrightarrow4\sqrt{2x-1}=12\Leftrightarrow\sqrt{2x-1}=3\Leftrightarrow\left\{{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

a)<=>\(\sqrt{4\left(x-2\right)}\)-2\(\sqrt{9\left(x-2\right)}\)+2\(\sqrt{16\left(x-2\right)}\)=12
<=>4\(\sqrt{x-2}\)=12

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

<=>x-2=9

<=>x=11

mik lm hơi tắt thông cảm

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

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

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

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

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

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

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

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

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

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

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

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

b,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}-16\sqrt{x+1}=0\) (dk \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}\left(4-3+2-16\right)=0\)

\(\Leftrightarrow\sqrt{x+1}.-13=0\)

\(\Leftrightarrow x=-1\)

\(\sqrt{16.\left(1-2x\right)}\) -\(\sqrt{4.3}\) =\(\sqrt{3x}\) +\(\sqrt{9.\left(1-2x\right)}\) 

(=)4\(\sqrt{1-2x}\) -2\(\sqrt{3x}\) =\(\sqrt{3x}\) +3\(\sqrt{1-2x}\) 

(=)\(\sqrt{1-2x}\) -3\(\sqrt{3x}\) =0

(=)1-2x=3\(\sqrt{3x}\) (=)1-2x=9.3x(=)1-2x=27x(=)29x=1(=)x=\(\frac{1}{29}\)