1. Đợi chút t tìm cách ngắn gọn.

2. ĐK: \(\left\{{}\begin{matrix}2x^2+8x+6\ge0\\x^2-1\ge0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge1\\x=-1\end{matrix}\right.\) (*)

BPT\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\3x^2+8x+5+2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\le\left(2x+2\right)^2\left(1\right)\end{matrix}\right.\)

Giải (1) \(\Leftrightarrow x^2-1-2\sqrt{\left(2x^2+8x+6\right)\left(x^2-1\right)}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\left(\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\right)\ge0\)

TH1: \(\sqrt{x^2-1}=0\Leftrightarrow x=\pm1\) (tm)

TH2: \(x^2-1\ne0\)

\(\Leftrightarrow\sqrt{x^2-1}-2\sqrt{2x^2+8x+6}\ge0\)

\(\Leftrightarrow\sqrt{x^2-1}\ge2\sqrt{2x^2+8x+6}\)

\(\Leftrightarrow x^2-1\ge8x^2+32x+24\)

\(\Leftrightarrow7x^2+32x+25\le0\)

\(\Leftrightarrow-\frac{25}{7}\le x\le-1\) kết hợp đk (*) và đk để giải bpt

=>\(x=-1\)

Vậy \(x=\pm1\)

3. ĐK: \(x\ge\frac{4}{5}\)

\(BPT\Leftrightarrow\sqrt{5x-4}-\sqrt{3x-2}+\sqrt{4x-3}-\sqrt{2x-1}>0\)

\(\Leftrightarrow\frac{2x-2}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{2x-2}{\sqrt{4x-3}+\sqrt{2x-1}}>0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{5x-4}+\sqrt{3x-2}}+\frac{1}{\sqrt{4x-3}+\sqrt{2x-1}}\right)>0\)

\(\Leftrightarrow x-1>0\) \(\Leftrightarrow x>1\)

Vậy \(x>1\)

a/ Đặt \(\sqrt{x^2-3x+5}=t>0\)

\(\Leftrightarrow t^2-5-t>1\Leftrightarrow t^2-t-6>0\)

\(\Rightarrow\left[{}\begin{matrix}t>3\\t< -2\left(l\right)\end{matrix}\right.\) \(\Rightarrow\sqrt{x^2-3x+5}>3\)

\(\Leftrightarrow x^2-3x+5>9\Leftrightarrow x^2-3x-4>0\Rightarrow\left[{}\begin{matrix}x>4\\x< -1\end{matrix}\right.\)

b/ ĐKXĐ: \(x\ge1\)

Đặt \(\sqrt[4]{x-\sqrt{x^2-1}}=t>0\Rightarrow\sqrt[4]{x+\sqrt{x^2-1}}=\frac{1}{t}\)

\(\Leftrightarrow t+\frac{4}{t^2}-3< 0\)

\(\Leftrightarrow t^3-3t^2+4< 0\)

\(\Leftrightarrow\left(t+1\right)\left(t-2\right)^2< 0\)

Do \(t>0\Rightarrow t+1>0\Rightarrow VT\ge0\Rightarrow\) BPT vô nghiệm

ĐKXĐ: \(x^2+x-1\ge0\)

\(\Rightarrow3x^2-x+1>3\sqrt{\left(x^2-x+1\right)\left(x^2+x-1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow2a^2+b^2>3ab\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}2a< b\\a>b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}< \sqrt{x^2+x-1}\\\sqrt{x^2-x+1}>\sqrt{x^2+x-1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)< x^2+x-1\\x^2-x+1>x^2+x-1\end{matrix}\right.\)

\(\Leftrightarrow...\) (nhớ kết hợp ĐKXĐ ban đầu)

ĐKXĐ: \(x^2\ge2\)

Đặt \(\sqrt{x^2-2}=a\ge0\)

BPT tương đương: \(\dfrac{1}{\sqrt{a^2+3}}+\dfrac{1}{\sqrt{3a^2+11}}\le\dfrac{2}{a+1}\)

Ta có: \(VT^2\le2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+11}\right)< 2\left(\dfrac{1}{a^2+3}+\dfrac{1}{3a^2+1}\right)=\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\)

Mặt khác ta có: \(\left(a-1\right)^4\ge0\Leftrightarrow a^4-4a^3+6a^2-4a+1\ge0\)

\(\Leftrightarrow3a^4+10a^2+3\ge2a^4+4a^3+4a^2+4a+2\)

\(\Leftrightarrow\left(3a^2+1\right)\left(a^2+3\right)\ge2\left(a^2+1\right)\left(a+1\right)^2\)

\(\Rightarrow\dfrac{8\left(a^2+1\right)}{\left(3a^2+1\right)\left(a^2+3\right)}\le\dfrac{4}{\left(a+1\right)^2}\)

\(\Rightarrow VT^2< \dfrac{4}{\left(a+1\right)^2}\Rightarrow VT< \dfrac{2}{a+1}\)

\(\Rightarrow\) BPT đã cho đúng với mọi \(a\ge0\) hay nghiệm của BPT là \(x^2\ge2\)

ĐKXĐ: \(x\ge\frac{2}{3}\)

\(\Leftrightarrow\sqrt{x+2}-2+x-\sqrt{3x-2}+x^2-2x\le0\)

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

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

\(\Leftrightarrow x-2\le0\Rightarrow x\le2\)

Vậy nghiệm của BPT là: \(\frac{2}{3}\le x\le2\)