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 \(x>1\)

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

- Với \(x=-\dfrac{3}{2}\) là nghiệm của BPT

- Với \(x>-\dfrac{3}{2}\Rightarrow2x+3>0\)

\(\Rightarrow\dfrac{3\left(2x-3\right)\left(2x+3\right)}{\sqrt{3x^2-3}}\le2x+3\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{\sqrt{3x^2-3}}\le1\)

\(\Rightarrow3\left(2x-3\right)\le\sqrt{3x^2-3}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3< 0\\\left\{{}\begin{matrix}2x-3\ge0\\9\left(2x-3\right)^2\le3x^2-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\left[{}\begin{matrix}x\ge\dfrac{3}{2}\\11x^2-36x+28\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{14}{11}\le x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< \dfrac{3}{2}\\\dfrac{3}{2}\le x\le2\end{matrix}\right.\) \(\Rightarrow-\dfrac{3}{2}< x\le2\)

Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}-\dfrac{3}{2}< x< -1\\1< x\le2\end{matrix}\right.\)

- Với \(x< -\dfrac{3}{2}\Rightarrow2x+3< 0\)

\(\dfrac{3\left(2x-3\right)\left(2x+3\right)}{\sqrt{3x^2-3}}\le2x+3\Leftrightarrow\dfrac{3\left(2x-3\right)}{\sqrt{3x^2-3}}\ge1\)

\(\Rightarrow3\left(2x-3\right)\ge\sqrt{3x^2-3}\)

Do \(x< -\dfrac{3}{2}\Rightarrow3\left(2x-3\right)< 0\Rightarrow\) BPT vô nghiệm

Vậy nghiệm của BPT là \(\left[{}\begin{matrix}-\dfrac{3}{2}\le x< -1\\1< x\le2\end{matrix}\right.\)

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 \(x=1\)

\(BPT\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(3x^2+2x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(2+\sqrt{x^2-2x+5}\right)\left(x+1\right)+\frac{2x\left(x+1\right)\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\le0\)

\(\Leftrightarrow\left(x+1\right)\text{[}2+\sqrt{x^2-2x+5}+\frac{2x\left(3x-1\right)}{2\sqrt{x^2+1}+\sqrt{x^2-2x+5}}\text{]}\le0\)

\(\Leftrightarrow\left(x+1\right)\left(4\sqrt{x^2+1}+2\sqrt{x^2-2x+5}+2\sqrt{\left(x^2+1\right)\left(x^2-2x+5\right)}+7x^2-4x+5\right)\)\(\le0\Leftrightarrow x+1\le0\Leftrightarrow x\le-1\)

ĐKXĐ: \(-1\le x\le2\)

\(\Leftrightarrow2x^2-2x-1+\sqrt{\left(x+1\right)\left(2-x\right)}\le0\)

Đặt \(\sqrt{\left(x+1\right)\left(2-x\right)}=t\ge0\)

\(\Rightarrow2x^2-2x=4-2t^2\)

BPT trở thành:

\(4-2t^2-1+t\le0\Leftrightarrow-2t^2+t+3\le0\Rightarrow\left[{}\begin{matrix}t\le-1\left(l\right)\\t\ge\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left(x+1\right)\left(2-x\right)\ge\frac{9}{4}\)

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

Vậy BPT có nghiệm duy nhất \(x=\frac{1}{2}\)

ĐKXĐ: \(x\ge3\)

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

\(\Leftrightarrow x-1>2x-5+2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow4-x>2\sqrt{x^2-5x+6}\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\\left(4-x\right)^2>4\left(x^2-5x+6\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\3x^2-12x+8< 0\end{matrix}\right.\)

\(\Rightarrow\dfrac{6-2\sqrt{3}}{3}< x< \dfrac{6+2\sqrt{3}}{3}\)

Kết hợp ĐKXĐ \(\Rightarrow3\le x< \dfrac{6+2\sqrt{3}}{3}\)

1) ĐKXĐ: \(\left[{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)

ta có: (-6).\(\sqrt{6x^2-18x+12}\) > \(6x^2-18x-60\)

⇔ \(6x^2-18x+12\) + \(2.3.\sqrt{6x^2-18x+12}+9-81\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+3\right)^2-9^2\) > 0

⇔ \(\left(\sqrt{6x^2-18x+12}+12\right).\left(\sqrt{6x^2-18x+12}-6\right)\) > 0

⇔ \(\sqrt{6x^2-18x+12}-6\) > 0

⇔ \(\sqrt{6x^2-18x+12}>6\)

⇔\(6x^2-18x+12>36\)

⇔ \(6x^2-18x-24>0\)

⇔\(\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\)

đối chiếu ĐKXĐ ban đầu ta được: x ϵ (-∞;-1) \(\cup\)(4;+∞)

b) ĐKXĐ: \(\forall x\) ϵ R

\(\left(x-2\right)\sqrt{x^2+4}-\left(x-2\right)\left(x+2\right)\le0\)

⇔\(\left(x-2\right)\left(\sqrt{x^2+4}-x-2\right)\le0\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\\sqrt{x^2+4}-x-2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\\sqrt{x^2+4}-x-2\ge0\end{matrix}\right.\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x^2+4\le x^2+4x+4\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x^2+4\ge x^2+4x+4\end{matrix}\right.\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le0\end{matrix}\right.\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)

Đối chiếu ĐKXĐ ta được x ϵ ( -∞;0) \(\cup\)( 2; +∞)