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 \(x^2\ge2\)

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; +∞)

ĐKXĐ: \(-2\le x\le3\)

\(\Leftrightarrow3x^3+3x^2-12x-12+x+4-3\sqrt{x+2}+5-x-3\sqrt{3-x}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(3x+6\right)+\frac{x^2-x-2}{x+4+3\sqrt{x+2}}+\frac{x^2-x-2}{5-x+3\sqrt{3-x}}\ge0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left[3\left(x+2\right)+\frac{1}{x+4+3\sqrt{x+2}}+\frac{1}{5-x+3\sqrt{3-x}}\right]\ge0\)

\(\Leftrightarrow x^2-x-2\ge0\)

\(\Rightarrow\left[{}\begin{matrix}-2\le x\le-1\\2\le x\le3\end{matrix}\right.\)

ĐKXĐ: \(-2\le x\le3\)

Do trên \(\left[-2;3\right]\) cả \(2x+5\) và \(x+4\) đều dương nên BPT tương đương:

\(\frac{1}{2x+5}\le\frac{1}{x+4}\Leftrightarrow x+4\le2x+5\Leftrightarrow x\ge-1\)

Vậy nghiệm của BPT là \(\left[{}\begin{matrix}x=-2\\-1\le x\le3\end{matrix}\right.\)