ĐKXĐ: \(x\ge-1\)

\(pt\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}=20\)

\(\Leftrightarrow4\sqrt{x+1}=20\Leftrightarrow\sqrt{x+1}=5\)

\(\Leftrightarrow x+1=25\Leftrightarrow x=24\left(tm\right)\)

\(\sqrt{9x+9}=20-\sqrt{x+1}\)

\(\Leftrightarrow x+1=25\)

hay x=24

\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

a, ĐK: \(x\in R\)

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

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

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

TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)

b, ĐK: \(x\ge-1\)

\(\sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\text{​​}\right)\)

\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}-2\sqrt{x+1}=2\left(x+1\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(\sqrt{x^2-x+1}+\sqrt{x^2-9x+9}=2x\)

=>\(\sqrt{x^2-x+1}-x+\sqrt{x^2-9x+9}-x=0\)

=>\(\dfrac{x^2-x+1-x^2}{\sqrt{x^2-x+1}+x}+\dfrac{x^2-9x+9-x^2}{\sqrt{x^2-9x+9}+x}=0\)

=>\(\left(-x+1\right)\left(\dfrac{1}{\sqrt{x^2-x+1}+x}+\dfrac{9}{\sqrt{x^2-9x+9}+x}\right)=0\)

=>-x+1=0

=>x=1

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

=>4x-4=1

hay x=5/4

\(\dfrac{1}{3}\sqrt[]{9x+9}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4x+4}{25}}=11\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt[]{9\left(x+1\right)}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4\left(x+1\right)}{25}}=11\)

\(\Leftrightarrow\sqrt[]{x+1}-2\sqrt[]{x+1}+\dfrac{16}{5}\sqrt[]{x+1}=11\)

\(\Leftrightarrow\dfrac{11}{5}\sqrt[]{x+1}=11\)

\(\Leftrightarrow\sqrt[]{x+1}=5\)

\(\Leftrightarrow x+1=25\)

\(\Leftrightarrow x=24\)

Ta có: \(\sqrt{25x-125}-3\cdot\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)

\(\Leftrightarrow5\sqrt{x-5}-3\cdot\dfrac{\sqrt{x-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\)

\(\Leftrightarrow3\sqrt{x-5}=6\)

\(\Leftrightarrow x-5=4\)

hay x=9

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: \(x\ge-1\))

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

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

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

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

\(\Leftrightarrow x+1=16\)

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

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

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

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)