Lời giải:

ĐKXĐ: $x\geq 2$

BPT $\Leftrightarrow x+1< 2x-3+2\sqrt{(x-1)(x-2)}$

$\Leftrightarrow 4-x< 2\sqrt{(x-1)(x-2)}$

$\Rightarrow (4-x)^2< 4(x-1)(x-2)$

$\Leftrightarrow 3x^2-4x-8>0$

$\Leftrightarrow x>\frac{2+2\sqrt{7}}{3}$ hoặc $x< \frac{2-2\sqrt{7}}{3}$

Kết hợp ĐKXĐ: suy ra $x> \frac{2+2\sqrt{7}}{3}$

ĐK: \(x\ge4;x\le0\)

TH1: \(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\Rightarrow bpt\) đúng

TH2: \(x\ne0;x\ne4\)

Bất phương trình tương đương:

\(\dfrac{x^2-3x+2}{x-3}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{x-3}\ge0\)

Lập bảng xét dấu:

Dựa vào bảng xét dấu, bất phương trình có nghiệm \(x\in\left[1;2\right]\cup\left(3;+\infty\right)\)

Kết luận: Bất phương trình đã cho có tập nghiệm \(x\in\left[1;2\right]\cup\left(3;+\infty\right)\cup\left\{0\right\}\)

bình phương hai vế ta có

\(\frac{-1}{x-1}

ĐKXĐ: \(x>\dfrac{3}{2}\)

\(\Leftrightarrow3-\left(2x-3\right)>2\sqrt{2x-3}\)

\(\Leftrightarrow3-x>\sqrt{2x-3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x^2-6x+9>2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x^2-8x+12\ge0\end{matrix}\right.\) \(\Rightarrow x\le2\)

Kết hợp ĐKXĐ \(\Rightarrow\dfrac{3}{2}< x\le2\)

giúp mình với ạ

1: Ta có: \(\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

\(=\dfrac{-\sqrt{x}}{x+\sqrt{x}+1}\)

2) Ta có: \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

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

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

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)