ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

\(\Leftrightarrow x^2-3x+2=0\)

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

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

1. ĐKXĐ: $x\geq \frac{-3}{5}$

PT $\Leftrightarrow 5x+3=3-\sqrt{2}$

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

2. ĐKXĐ: $x\geq \sqrt{7}$ 

PT $\Leftrightarrow (\sqrt{x}-7)(\sqrt{x}+7)=4$

$\Leftrightarrow x-49=4$

$\Leftrightarrow x=53$ (thỏa mãn)

\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)

\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)

\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)

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

=>\(\sqrt{3x+1}-2+2\sqrt{x+3}-4=3\sqrt{5x-1}-6\)

=>\(\dfrac{3x+1-4}{\sqrt{3x+1}+2}+2\left(\sqrt{x+3}-2\right)-3\left(\sqrt{5x-1}-2\right)=0\)

=>\(\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}+2\cdot\dfrac{x+3-4}{\sqrt{x+3}+2}-3\cdot\dfrac{5x-1-4}{\sqrt{5x-1}+2}=0\)

=>\(\left(x-1\right)\left(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{2}{\sqrt{x+3}+2}-\dfrac{15}{\sqrt{5x-1}+2}\right)=0\)

=>x-1=0

=>x=1

a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)

`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`

`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\) 

`b)sqrt{x-3}/sqrt{2x+1}=2`

ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)

`<=>x>=3`

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

`<=>x-3=8x+4`

`<=>7x=7`

`<=>x=1(l)`

`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`

`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`

`<=>|x-1|+|x-2|=3`

`**x>=2`

`pt<=>x-1+x-2=3`

`<=>2x=6`

`<=>x=3(tm)`

`**x<=1`

`pt<=>1-x+2-x=3`

`<=>3-x=3`

`<=>x=0(tm)`

`**1<=x<=2`

`pt<=>x-1+2-x=3`

`<=>=-1=3` vô lý

Vậy `S={0,3}`

\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-2\sqrt{2x^2+5x-3}=1+x\sqrt{2x-1}-2x\sqrt{x+3}\\ \Leftrightarrow\left(2x-2\right)-\left(2\sqrt{2x^2+5x-3}-4\right)=\left(x\sqrt{2x-1}-x\right)-\left(2x\sqrt{x+3}-4x\right)-3x+3\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(2x^2+5x-7\right)}{\sqrt{2x^2+5x-3}+4}=\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}-\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}-3\left(x-1\right)\\ \Leftrightarrow2\left(x-1\right)-\dfrac{2\left(x-1\right)\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x\left(x-1\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+4x}+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\2-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}-\dfrac{2x}{\sqrt{2x-1}+2}+\dfrac{2x}{\sqrt{x+3}+4x}+3=0\left(1\right)\end{matrix}\right.\)

Với \(x\ge\dfrac{1}{2}\Leftrightarrow-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+4}>-\dfrac{2\cdot8}{4}=-4\)

\(-\dfrac{2x}{\sqrt{2x-1}+2}>-\dfrac{1}{2};\dfrac{2x}{\sqrt{x+3}+4x}>0\)

Do đó \(\left(1\right)>2-4-\dfrac{1}{2}+3=\dfrac{1}{2}>0\) nên (1) vô nghiệm

Vậy PT có nghiệm duy nhất \(x=1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x^2+3x+1}=a\\\sqrt[3]{5x+1}=b\end{matrix}\right.\)

\(\Rightarrow a+a^3-b^3=b\)

\(\Leftrightarrow a-b+\left(a-b\right)\left(a^2+ab+b^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt[3]{x^2+3x+1}=\sqrt[3]{5x+1}\)

\(\Leftrightarrow x^2+3x+1=5x+1\)

\(\Leftrightarrow...\)