a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\x\ne\left\{3;11\right\}\end{matrix}\right.\)

Đặt \(\sqrt{x-2}=t\ge0\)

\(\Rightarrow\frac{3}{t-1}\ge\frac{5}{t-3}\)

\(\Leftrightarrow\frac{3}{t-1}-\frac{5}{t-3}\ge0\)

\(\Leftrightarrow\frac{3t-9-5t+5}{\left(t-1\right)\left(t-3\right)}\ge0\)

\(\Leftrightarrow\frac{-2t-4}{\left(t-1\right)\left(t-3\right)}\ge0\)

\(\Leftrightarrow\frac{t+2}{\left(t-1\right)\left(t-3\right)}\le0\)

\(\Leftrightarrow1< t< 3\)

\(\Rightarrow1< \sqrt{x-2}< 3\)

\(\Leftrightarrow1< x-2< 9\Rightarrow3< x< 11\)

b/

ĐKXĐ: \(x\ge3\)

- Với \(x=3\) BPT thỏa mãn

- Với \(x>3\Rightarrow\sqrt{x-3}>0\) BPT tương đương

\(x-\frac{1}{2-x}\le0\Leftrightarrow x+\frac{1}{x-2}\le0\)

\(\Leftrightarrow\frac{x^2-2x+1}{x-2}\le0\)

\(\Leftrightarrow\frac{\left(x-1\right)^2}{x-2}\le0\Rightarrow\) không tồn tại x thỏa mãn

Vậy BPT có nghiệm duy nhất \(x=3\)

c/

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\4-x^2\ge0\\\sqrt{4-x^2}\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\-2\le x\le2\\x\ne\pm\sqrt{3}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le2\\x\ne\sqrt{3}\end{matrix}\right.\)

BPT tương đương:

\(\frac{2\left(\sqrt{x-1}-2\right)}{\sqrt{4-x^2}-1}+\sqrt{x-1}-2\ge0\)

\(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\frac{2}{\sqrt{4-x^2}-1}+1\right)\ge0\)

Do \(x\le2\Rightarrow\sqrt{x-1}\le1\Rightarrow\sqrt{x-1}-2< 0\)

BPt tương đương:

\(\frac{2}{\sqrt{4-x^2}-1}+1\le0\)

\(\Leftrightarrow\frac{1+\sqrt{4-x^2}}{\sqrt{4-x^2}-1}\le0\)

\(\Leftrightarrow\sqrt{4-x^2}-1< 0\) (do \(1+\sqrt{4-x^2}>0\) \(\forall x\))

\(\Leftrightarrow\sqrt{4-x^2}< 1\Leftrightarrow x^2>3\Rightarrow x>\sqrt{3}\)

Vậy nghiệm của BPT đã cho là: \(\sqrt{3}< x\le2\)

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

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

=>(x+5)(x-2)>=0

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

ĐK: \(x\ge1;x\le-2\)

\(\sqrt{x^2-1}+\sqrt{x^2-x}\le\sqrt{x^2+x-2}\)

\(\Leftrightarrow2x^2-x-1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le x^2+x-2\)

\(\Leftrightarrow x^2-2x+1+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

\(\Leftrightarrow\left(x-1\right)^2+2\sqrt{\left(x^2-1\right)\left(x^2-x\right)}\le0\)

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

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

Vậy bất phương trình có nghiệm \(x=1\)