b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)

Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)

1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)

\(3x+2x^2-6-4x-2x^2-10x-6=0\)

\(-11x=12\)

\(x=-\dfrac{12}{11}\)

2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\left(x-2\right)\left(x+2-x+5\right)=0\)

\(7\left(x-2\right)=0\)

\(\Leftrightarrow x=2\)

1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)

2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)

3, bạn xem lại đề 

5, đk x khác -4 ; 4 

\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)

\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)

\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm) 

3) 

\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\\ \Leftrightarrow\dfrac{2\left(x-3\right)}{6}-\dfrac{3\left(x+2\right)}{6}=\dfrac{x}{6}\\ \Leftrightarrow2x-6-3x-6=x\\ \Leftrightarrow2x-3x-x=6+6\\ \Leftrightarrow-2x=12\\ \Leftrightarrow x=-6\)

Vậy PT có tập nghiệm S = { -6 }

a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)

b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)

\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)

\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)

\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)

\(\Leftrightarrow x\left(6-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: S={0;6}

c) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)

\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)

\(\Leftrightarrow30-6x=6x-8\)

\(\Leftrightarrow30-6x-6x+8=0\)

\(\Leftrightarrow-12x+38=0\)

\(\Leftrightarrow-12x=-38\)

\(\Leftrightarrow x=\dfrac{19}{6}\)

Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)

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

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow6x+4-3x-1=12x+10\)

\(\Leftrightarrow3x+3-12x-10=0\)

\(\Leftrightarrow-9x-7=0\)

\(\Leftrightarrow-9x=7\)

\(\Leftrightarrow x=-\dfrac{7}{9}\)

Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)

2.a)

\(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\)

\(\Leftrightarrow-3x>6\)

\(\Leftrightarrow3>\dfrac{6}{-3}\)

\(\Leftrightarrow x< -2\)

Vậy nghiệm của bpt \(S=\left\{-2\right\}\)

2.b)

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

\(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\)

\(\Leftrightarrow4x+4-12\ge3x-6\)

\(\Leftrightarrow4x-3x\ge-6-4+12\)

\(\Leftrightarrow x\ge2\)

vậy nghiệm của bpt x\(\ge\)2

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

Dài quá c ơi :<

a) 4x -8 ≥ 3(3x-1)-2x +1

⇒4x -8 ≥7x -2

⇒4x -7x ≥ -2 +8

⇒-3x ≥ 6

⇒x≤-2

Vậy bpt có nghiệm là:{x|x≤-2}

b) (x-3)(x+2)+(x+4)²≤ 2x (x+5)+4

⇔ x²+2x - 3x - 6 +x² + 8x +16≤ 2x² + 10x +4

⇔ x² +2x - 3x + x² + 8x - 2x²- 10x ≤ 4+6-16

⇔ -3x ≤ -6

⇔ x≥ 2

Vậy bpt có tập nghiệm là: {x|x≥2}