2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

ĐKXĐ: x+3>=0 và 7-x>=0 và 2x-8>=0

=>x=-3 và x<=7 và x>=4

=>4<=x<=7

\(\sqrt{x+3} - \sqrt{7-x} - \sqrt{2x-8} > 0\)

=>\((\sqrt{x+3}-\sqrt{8})+(\sqrt{2}-\sqrt{7-x})+(\sqrt{2}-\sqrt{2x-8})>0\)

=>\(\frac{x-5}{\sqrt{x+3} + \sqrt{8}}+\frac{x-5}{\sqrt{2} + \sqrt{7-x}}-\frac{2(x-5)}{\sqrt{2} + \sqrt{2x-8}}>0\)

=>\((x-5)\left[\frac{1}{\sqrt{x+3} + 2\sqrt{2}}+\frac{1}{\sqrt{2} + \sqrt{7-x}}-\frac{2}{\sqrt{2} + \sqrt{2x-8}}\right]>0\)

Khi x∈(5;7] thì ta có:

\(\sqrt{x+3}+2\sqrt{2}>2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\)

=>\(\frac{1}{\sqrt{x+3} + 2\sqrt{2}}<\frac{1}{4\sqrt{2}}\) (2)

\(\sqrt{7-x}<\sqrt{2}\)

=>\(\frac{1}{\sqrt{2} + \sqrt{7-x}}>\frac{1}{2\sqrt{2}}\) (1)

\(\sqrt{2x-8}>\sqrt{2}\)

=>\(\frac{2}{\sqrt{2} + \sqrt{2x-8}}<\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\) (3)

Từ (1),(2),(3) suy ra \(\left[\frac{1}{\sqrt{x+3} + 2\sqrt{2}}+\frac{1}{\sqrt{2} + \sqrt{7-x}}-\frac{2}{\sqrt{2} + \sqrt{2x-8}}\right]>0\forall x\) ∈(5;7]

=>x-5>0

=>x>5

Vậy: x∈(5;7]

Bất phương trình : \(\Leftrightarrow2^{\frac{x+1}{2}}.2^{\frac{4x-2}{3}}.2^{9-3x}>2^{\frac{3}{2}}.2^{-3}\)

                            \(\Leftrightarrow2^{\frac{x+1}{2}+\frac{4x-2}{3}+9-3x}>2^{\frac{3}{2}-3}\)

                            \(\Leftrightarrow x< \frac{62}{7}\)

Vậy bất phương trình có tập nghiệm là \(S=\left(-\infty;\frac{62}{7}\right)\)

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

Dễ thấy: \(x^2+2x+2>0;x^2-2x+3>0\)

\(\Rightarrow bpt\Leftrightarrow\left(\sqrt{x^2+2x+2}\right)^2>\left(\sqrt{x^2-2x+3}\right)^2\)

\(\Leftrightarrow x^2+2x+2>x^2-2x+3\)

\(\Leftrightarrow4x>1\Leftrightarrow x>\frac{1}{4}\)

Vậy nghiệm của bpt là \(T=\left(\frac{1}{4};+\infty\right)\)

\(\Leftrightarrow\left(\sqrt[3]{x+1}-1\right)+\left(\sqrt{2x+4}-2\right)< -x\sqrt{2}\)

=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)

=>x<0

=>-1<x<0

ĐKXĐ: \(-\dfrac{3}{2}\le x\le4\)

BPT tương đương:

\(6+2\sqrt{\left(x+2\right)\left(4-x\right)}>2x+3\)

\(\Leftrightarrow2\sqrt{-x^2+2x+8}>2x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\4\left(-x^2+2x+8\right)>4x^2-12x+9\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\8x^2-20x-23< 0\end{matrix}\right.\end{matrix}\right.\) 

\(\Rightarrow-\dfrac{3}{2}\le x< \dfrac{5+\sqrt{71}}{4}\)