1) \(\dfrac{x+2\sqrt[]{x}}{\sqrt[]{x}-1}=8\left(1\right)\)

Điều kiện \(\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

\(\Leftrightarrow x-6\sqrt[]{x}+8=0\left(2\right)\)

Đặt \(t^2=x\Leftrightarrow t=\sqrt[]{x}\)

\(\left(2\right)\Leftrightarrow t^2-6t+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x}=2\\\sqrt[]{x}=4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\) (thỏa điều kiện)

2) \(\sqrt[]{\dfrac{2x-3}{x-1}}=2\left(1\right)\)

Điều kiện \(\dfrac{2x-3}{x-1}\ge0\Leftrightarrow\left[{}\begin{matrix}x< 1\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

\(\Leftrightarrow2x-3=4\left(x-1\right)\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) (thỏa điều kiện)

Bài 2: 

Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

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

b: Ta có: \(\sqrt{4x+8}+\dfrac{1}{3}\sqrt{9x+18}=3\sqrt{\dfrac{x+2}{4}}+\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}+\dfrac{1}{3}\cdot3\sqrt{x+2}-\dfrac{3}{2}\sqrt{x+2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}\cdot\dfrac{3}{2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{2\sqrt{2}}{3}\)

\(\Leftrightarrow x+2=\dfrac{8}{9}\)

hay \(x=-\dfrac{10}{9}\)

a) \(x+1+\dfrac{2}{x+3}=\dfrac{x+5}{x+3}\)

\(\Leftrightarrow x+\dfrac{x+5}{x+3}=\dfrac{x+5}{x+3}\)

\(\Leftrightarrow x=0\)

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

\(\Leftrightarrow x+x+\dfrac{3}{x-1}=\dfrac{3x}{x-1}\)

\(\Leftrightarrow x+\dfrac{x\left(x-1\right)+3}{x-1}=\dfrac{3x}{x-1}\)

\(\Leftrightarrow x+\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}\)

\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x}{x-1}-x\)

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

\(\Leftrightarrow\dfrac{x^2-x+3}{x-1}=\dfrac{3x-x^2+x}{x-1}\)

\(\Leftrightarrow x^2-x+3=3x-x^2+x\) ( điều kiện \(x\ne1\) )

\(\Leftrightarrow2x^2-5x+3=0\)

\(\Delta=b^2-4ac\)

\(\Delta=1\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3}{2}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=1\left(loại\right)\end{matrix}\right.\)

Vậy \(x=\dfrac{3}{2}\)

c) \(\dfrac{x^2-4x-2}{\sqrt{x-2}}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-4x-2=\sqrt{\left(x-2\right)^2}\) ( điều kiện \(x>2\) )

\(\Leftrightarrow x^2-4x-2=x-2\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=5\end{matrix}\right.\)

Vậy \(x=5\)

d) \(\dfrac{2x^2-x-3}{\sqrt{2x-3}}=\sqrt{2x-3}\)

\(\Leftrightarrow2x^2-x-3=\sqrt{\left(2x-3\right)^2}\) ( điều kiện \(x>\dfrac{3}{2}\) )

\(\Leftrightarrow2x^2-x-3=2x-3\)

\(\Leftrightarrow2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

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

Vậy phương trình vô nghiệm

1) \(\sqrt{x^2+1}=\sqrt{5}\)

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

\(\Leftrightarrow x^2=5-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x^2=2^2\)

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

2) \(\sqrt{2x-1}=\sqrt{3}\) (ĐK: \(x\ge\dfrac{1}{2}\)) 

\(\Leftrightarrow2x-1=3\)

\(\Leftrightarrow2x=3+1\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=\dfrac{4}{2}\)

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

3) \(\sqrt{43-x}=x-1\) (ĐK: \(x\le43\))

\(\Leftrightarrow43-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1=43-x\)

\(\Leftrightarrow x^2-x-42=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

4) \(x-\sqrt{4x-3}=2\) (ĐK: \(x\ge\dfrac{3}{4}\))

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

\(\Leftrightarrow4x-3=\left(x-2\right)^2\)

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

\(\Leftrightarrow x^2-8x+7=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

5) \(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\) (ĐK: \(x\ge0\))

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

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=3-2\)

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

\(\Leftrightarrow x=1^2\)

\(\Leftrightarrow x=1\left(tm\right)\)

1)

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

Vậy PT có nghiệm `x=2` hoặc `x=-2`

2)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}=\sqrt{3}\\ \Leftrightarrow2x-1=3\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy PT có nghiệm `x=2`

3)

\(ĐKXĐ:x\le43\)

PT trở thành:

\(43-x=\left(x-1\right)^2=x^2-2x+1\\ \Leftrightarrow43-x-x^2+2x-1=0\\ \Leftrightarrow-x^2+x+42=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm `x=-6` hoặc `x=7`

4)

ĐKXĐ: \(x\ge\dfrac{3}{4}\)

PT trở thành:

\(\sqrt{4x-3}=x-2\\ \Leftrightarrow4x-3=\left(x-2\right)^2=x^2-4x+4\\ \Leftrightarrow4x-3-x^2+4x-4=0\\ \Leftrightarrow-x^2+8x-7=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=7\left(tm\right)\end{matrix}\right.\)

Vậy PT có nghiệm \(x=1\) hoặc \(x=7\)

5) 

ĐKXĐ: \(x\ge0\)

PT trở thành:

\(\sqrt{x+3}=2\sqrt{x}+2\\ \Leftrightarrow x+3=\left(2\sqrt{x}+2\right)^2=4x+8\sqrt{x}+4\\ \Leftrightarrow x+3-4x-8\sqrt{x}-4=0\\ \Leftrightarrow-3x-8\sqrt{x}-1=0\left(1\right)\)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)

Khi đó:

(1)\(\Leftrightarrow3t^2+8t+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-4+\sqrt{13}}{3}\left(loại\right)\\t=\dfrac{-4-\sqrt{13}}{3}\left(loại\right)\end{matrix}\right.\)

Vậy PT vô nghiệm.

Bài 1:

$\sqrt{x^2+1}=\sqrt{5}$

$\Leftrightarrow x^2+1=5$

$\Leftrightarrow x^2-4=0$

$\Leftrightarrow (x-2)(x+2)=0$

$\Leftrightarrow x-2=0$ hoặc $x+2=0$

$\Leftrightarrow x=\pm 2$ (đều tm)

2. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow 2x-1=3$

$\Leftrightarrow 2x=4$

$\Leftrightarrow x=2$ (tm) 

3. ĐKXĐ: $x\leq 43$

PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 43-x=(x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-42=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+6)(x-7)=0\end{matrix}\right.\)

$\Rightarrow x=7$ (tm) 

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}`