ĐKXĐ: \(x\ge3\)

Khi đó \(\sqrt{2x-1}\ge\sqrt{5}>1\Rightarrow\sqrt{2x-1}-1>0\)

Đồng thời \(\sqrt{x+3}>\sqrt{x-3}\) \(\forall x\Rightarrow\sqrt{x+3}-\sqrt{x-3}>0\)

Do đó BPT tương đương:

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

\(\Leftrightarrow\sqrt{x^2-9}-x+3\ge\sqrt{2x-1}-1\)

\(\Leftrightarrow\sqrt{x^2-9}\ge x-4+\sqrt{2x-1}\)

Do \(x-4+\sqrt{2x-1}\ge3-4+\sqrt{5}>0;\forall x\ge3\) nên BPT tương đương:

\(x^2-9\ge x^2-8x+16+2x-1+2\left(x-4\right)\sqrt{2x-1}\)

\(\Leftrightarrow\left(x-4\right)\sqrt{2x-1}-3\left(x-4\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(\sqrt{2x-1}-3\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(\frac{2x-1-9}{\sqrt{2x-1}+3}\right)\le0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)\le0\Leftrightarrow4\le x\le5\)

ttiiok

a,\(2x\left(x-3\right)=x-3.\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy ..... 

b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)

\(\Leftrightarrow x^2+2x-5x+10=8\)

\(\Leftrightarrow x^2-3x+10-8=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)

Vậy ....

\(\frac{2x+1}{4}-\frac{x-5}{3}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{\left(2x+1\right)\cdot3}{4\cdot3}-\frac{\left(x-5\right)\cdot4}{3\cdot4}< \frac{4x-1}{12}+12.\)

\(\Leftrightarrow\frac{6x+3}{12}-\frac{4x-20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{6x+3-4x+20}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23}{12}< \frac{4x-1}{12}+12\)

\(\Leftrightarrow\frac{2x+23-4x+1}{12}< 12\)

\(\Leftrightarrow\frac{-2x+24}{12}< 12\)

\(\Leftrightarrow-2x+24< 144\)

\(\Leftrightarrow-2x< 120\)

\(\Leftrightarrow x< -60\)

\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}+\frac{1}{x^2+2x-3}=1.\)

\(ĐK:\hept{\begin{cases}x-1\ne0\\x+3\ne\\x^2+2x-3\ne0\end{cases}0}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\Leftrightarrow-3\end{cases}}\)

\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)+4-x^2-2x+3=0\)

\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5+4-x^2-2x+3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\Leftrightarrow x=-3\) (loại)

 Vậy pt vô N^o

Số t tính đc rất thú dị :) 

a/ ĐKXĐ: ....

\(VT=\sqrt{11+x}+\sqrt{1-x}\ge\sqrt{11+x+1-x}=\sqrt{12}\)

\(VP=2-\frac{x^2}{4}\le2< \sqrt{12}\)

\(\Rightarrow VP< VT\Rightarrow\) BPT vô nghiệm

b/

ĐKXĐ: ...

- Với \(x\le0\Rightarrow VT\le0< VP\Rightarrow\) BPT vô nghiệm

- Với \(x>0\) \(\Rightarrow x>2\) hai vế đều dương, bình phương:

\(x^2+\frac{4x^2}{x^2-4}+\frac{4x^2}{\sqrt{x^2-4}}>45\)

\(\Leftrightarrow\frac{x^4}{x^2-4}+\frac{4x^2}{\sqrt{x^2-4}}-45>0\)

Đặt \(\frac{x^2}{\sqrt{x^2-4}}=t>0\)

\(\Rightarrow t^2+4t-45>0\Rightarrow\left[{}\begin{matrix}t< -9\left(l\right)\\t>5\end{matrix}\right.\)

\(\Rightarrow\frac{x^2}{\sqrt{x^2-4}}>5\Leftrightarrow x^4>25\left(x^2-4\right)\)

\(\Leftrightarrow x^4-25x^2+100>0\Rightarrow\left[{}\begin{matrix}x^2< 5\\x^2>20\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2< x< \sqrt{5}\\x>2\sqrt{5}\end{matrix}\right.\)

c/

ĐKXĐ: \(-2\le x\le2\)

Do \(-2\le x\le2\Rightarrow x+2\ge0\Rightarrow VT\ge0\) \(\forall x\)

Mà \(VP=-2x-8=-2\left(x+2\right)-4\le-4< 0\)

\(\Rightarrow VP< VT\)

Vậy BPT đã cho vô nghiệm