ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)

\(\dfrac{1}{x+2}< \dfrac{3}{x-3}\\ \Leftrightarrow x-3< 3\left(x+2\right)\\ \Leftrightarrow x-3< 3x+6\\ \Leftrightarrow2x+9< 0\\ \Rightarrow x< -\dfrac{9}{2}\)

\(\dfrac{15x-2}{4}-\dfrac{x^2+1}{3}>\dfrac{x\left(1-2x\right)}{6}+\dfrac{x-3}{2}\\ \Leftrightarrow3\left(15x-2\right)-4\left(x^2+1\right)>2x\left(1-2x\right)+6\left(x-3\right)\\ \Leftrightarrow45x-6-4x^2-4>2x-4x^2+6x-18\\ \Leftrightarrow45x-6x-2x>6+4-18\\ \Leftrightarrow37x>-8\\ \Leftrightarrow x>-\dfrac{8}{37}\)

\(\dfrac{3\left(15x-2\right)}{12}-\dfrac{4\left(x^2+1\right)}{12}>\dfrac{2x\left(1-2x\right)}{12}+\dfrac{6\left(x-3\right)}{12}\)

\(45x-6-\left(4x^2+4\right)>2x-4x^2+6x-18\)

\(45x-4x^2+4x^2-2x-6x>6+4-18\)

\(37x>-8\)

\(x>\dfrac{-8}{37}\)

\(\Leftrightarrow\dfrac{2}{x^2-3x+2}-\dfrac{3}{x^2+5x+4}\ge0\)

\(\Leftrightarrow\dfrac{-x^2+19x+2}{\left(x^2-3x+2\right)\left(x^2+5x+4\right)}\ge0\)

\(\Leftrightarrow\dfrac{-x^2+19x+2}{\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}2< x\le\dfrac{19+3\sqrt{41}}{2}\\\dfrac{19-3\sqrt{41}}{2}\le x< 1\\-4< x< -1\end{matrix}\right.\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

c: 2(x-1)+x=0

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)

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

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

- TH1: Với \(x>1\) BPT tương đương:

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

\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)

Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)

BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)

\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)

TH2: với \(x< 1\) BPT tương đương:

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

\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)

Tương tự như trên, đặt  \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)

\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)

Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

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/

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

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

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

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

$\Leftrightarrow x+1>0$

$\Leftrightarrow x>-1$