ĐKXĐ : \(\left\{{}\begin{matrix}2x-1>0\\y+2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>-2\end{matrix}\right.\)

PT ( I ) \(\Leftrightarrow\left(\sqrt{\dfrac{2x-1}{y+2}}+\sqrt{\dfrac{y+2}{2x-1}}\right)^2=4\)

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

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

Từ PT ( II ) ta được : \(x=12-y\)

- Thế x vào PT trên ta được : \(\dfrac{2\left(12-y\right)}{y+2}+\dfrac{y+2}{2\left(12-y\right)}=2\)

\(\Leftrightarrow4\left(y-12\right)^2+\left(y+2\right)^2=4\left(12-y\right)\left(y+2\right)\)

\(\Leftrightarrow4\left(y^2-24y+144\right)+y^2+4y+4=4\left(12y+24-y^2-2y\right)\)

\(\Leftrightarrow4y^2-96y+576+y^2+4y+4-40y-96+4y^2=0\)

\(\Leftrightarrow9y^2-132y+484=0\)

\(\Leftrightarrow y=\dfrac{22}{3}\left(TM\right)\)

- Thay lại vào PT ta được : \(x=\dfrac{14}{3}\)

Vậy phương trình có nghiệm là \(S=\left\{\left(\dfrac{22}{3};\dfrac{14}{3}\right);\left(\dfrac{14}{3};\dfrac{22}{3}\right)\right\}\)

\(ĐKXĐ:\left[{}\begin{matrix}x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\)

Đặt \(\sqrt{\dfrac{2x+1}{x-1}}=a\ge0\Rightarrow\dfrac{2x+1}{x-1}=a^2\)

Phương trình trở thành :

\(a^2+a-3=0\)

\(\text{Δ}=1^2-4\cdot1\cdot\left(-3\right)=13>0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{13}}{2}\left(t.m\right)\\x=\dfrac{-1-\sqrt{13}}{2}\left(loai.vi.a\ge0\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{2x+1}{x-1}=\dfrac{7-\sqrt{13}}{2}\Leftrightarrow2+\dfrac{3}{x-1}=2+\dfrac{3-\sqrt{13}}{2}\\ \Leftrightarrow\dfrac{3}{x-1}=\dfrac{3-\sqrt{13}}{2}\Leftrightarrow x-1=\dfrac{6}{3-\sqrt{13}}\\ \Leftrightarrow x-1=\dfrac{-9-3\sqrt{13}}{2}\Leftrightarrow x=\dfrac{-7-3\sqrt{13}}{2}\)

a.

ĐKXĐ: \(x>0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)

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)