\(P=\dfrac{2+x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-1}\\ P=\dfrac{\left(2-\sqrt{x}\right)\left(x+\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)^2}\)

ptr thiếu 1 vế rồi. hay là rút gọn nhỉ?

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

làm lại nhé :((( 

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

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) 

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

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

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

- TH1: Với \(x>1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

ĐKXĐ: `{(x+1>0),(x ne0):} <=> {(x> -1),(x ne 0):}`

`2/(sqrt(x+1))+1/(x sqrt(x+1)) =1/x`

`<=>(2x+1)/(x sqrt(x+1)) =1/x`

`<=>x(2x+1)=x sqrt(x+1)`

`<=>2x+1=sqrt(x+1)`

`=>(2x+1)^2=x+1`

`<=>4x^2+4x+1=x+1`

`<=>4x^2+3x=0`

`<=>x(4x+3)=0`

`<=>[(x=0\ (KTM)),(x=-3/4):}`

Thay `x=-3/4` vào PT ban đầu `=>` Không thỏa mãn.

Vậy phương trình vô nghiệm.

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)

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

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

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

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

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

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

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