2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

Lời giải:

b/

\(\frac{3x+5}{2x^2-5x+3}\geq 0\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} 3x+5\geq 0\\ 2x^2-5x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x+5\leq 0\\ 2x^2-5x+3<0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq \frac{-5}{3}\\ x>\frac{3}{2}(\text{hoặc}) x< 1\end{matrix}\right.\\ \left\{\begin{matrix} x\leq \frac{-5}{3}\\ 1< x< \frac{3}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x>\frac{3}{2}\\ \frac{-5}{3}\leq x< 1\end{matrix}\right.\ \)

c/

$2x^3+x+3>0$

$\Leftrightarrow 2x^2(x+1)-2x(x+1)+3(x+1)>0$

$\Leftrightarrow (x+1)(2x^2-2x+3)>0$

$\Leftrightarrow (x+1)[x^2+(x-1)^2+2]>0$

$\Leftrightarrow x+1>0$

$\Leftrightarrow x>-1$

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

\(\Leftrightarrow4\left(5x^2-3\right)+5\left(3x-1\right)< 10x\left(x+3\right)-100\)

\(\Leftrightarrow20x^2-12+15x-5< 10x^2+30x-100\)

\(\Leftrightarrow10x^2-15x+83< 0\)

\(\Leftrightarrow10\left(x-\frac{3}{4}\right)^2+\frac{619}{8}< 0\)

Bất phương trình vô nghiệm

a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)

\(\Leftrightarrow6x+2-20+8x>8x-6-6\)

\(\Leftrightarrow14x-18-8x+12>0\)

\(\Leftrightarrow6x-6>0\)

\(\Leftrightarrow6x>6\)

hay x>1

Vậy: S={x|x>1}

b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)

\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)

\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)

\(\Leftrightarrow-1< 0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

\(BPT\Leftrightarrow\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\ge\dfrac{\left(3x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^2-x-2x+2-3x^2-3x-2x-2-2x^2-2\ge0\)

\(\Leftrightarrow-4x^2-8x-2\ge0\)

\(\Leftrightarrow x^2+2x+\dfrac{1}{2}\ge0\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{1}{2}\ge0\)

Vậy bất phương trình luôn đúng \(\forall x\).

ĐKXĐ: \(x\ne1,-1\)

Ta có: \(\dfrac{x-2}{x+1}\ge\dfrac{3x+2}{x-1}-2\)

\(\dfrac{x-2}{x+1}\ge\dfrac{3x+2-2\left(x-1\right)}{x-1}\)

\(\dfrac{x-2}{x+1}-\dfrac{3x+2-2x+2}{x-1}\ge0\)

\(\dfrac{x-2}{x+1}-\dfrac{x+4}{x-1}\ge0\)

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

\(\dfrac{x^2-3x+2-x^2+3x+4}{x^2-1}\ge0\)

\(\dfrac{6}{x^2-1}\ge0\)

\(\Rightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x>1\end{matrix}\right.\)(TM)

Biểu thức vế trái có nghĩa khi 

 \(x\ne-2;x\ne1\\ \dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}\Leftrightarrow\dfrac{x-2}{x+2}+\dfrac{x+1}{x-1}-2>0\\ \dfrac{\left(x-2\right)\left(x-1\right)+\left(x+2\right)\left(x+1\right)-2\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\\ >0\\ \Leftrightarrow\dfrac{8-2x}{\left(x+2\right)\left(x-1\right)}>0\\ \Leftrightarrow\dfrac{4-x}{\left(x+2\right)\left(x-1\right)}>0\)  

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x-4\right)< 0\) 

Lập bảng xét dấu

Vậy nghiệm của bất pt là 

\(x< -2.hay.1< x< 4\) 

Ta có: \(\dfrac{x-1}{3}-\dfrac{3x+5}{2}\ge1-\dfrac{4x+5}{6}\)

\(\Leftrightarrow2\left(x-1\right)-3\left(3x+5\right)\ge6-4x-5\)

\(\Leftrightarrow2x-2-9x-15-6+4x+5\ge0\)

\(\Leftrightarrow-3x\ge18\)

hay \(x\le-6\)

=>\(\dfrac{3x+5}{3}>\dfrac{-4}{6x+5}\)

=>\(\dfrac{18x^2+15x+30x+25+12}{3\left(6x+5\right)}>0\)

=>\(\dfrac{18x^2+45x+37}{3\left(6x+5\right)}>0\)

=>6x+5>0

=>x>-5/6