a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x²+2)

<=> x²+3x+2+x²-3x+2 = 2x²+4

<=> x²+x²-2x²+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

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

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)²-4(x²-2x+1) = 0

<=> x²+2x+1-4x²+8x-4 = 0

<=> -3x²+10x-3 = 0

giải phương trình

b) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)

<=> \(\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{1\left(x-2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

<=> x²+2x-x+2=2

<=> x²+x=2-2

<=> x²+x=0

<=>x(x+1)=0

<=>x=0 hoặc x+1=0

<=>x=0 hoặc x = -1

a) \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)

<=>\(\frac{1.x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

<=> x-3 =10x-15

<=> x-10x= -15+3

<=> -9x = -12

<=> x = \(\frac{-12}{-9}\)

<=> x = \(\frac{4}{3}\)

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

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

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+x=2-2\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

KL ::

\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)    ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x+2\right)\left(x-2\right)}\)

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

\(\Leftrightarrow0x=0\)

KL : PT vô số nghiệm 

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

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

\(\Leftrightarrow-2x=-2\)

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

\(3x^2+7x-20=0\)

Ta có \(\Delta=7^2+4.3.20=289,\sqrt{\Delta}=17\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-7+17}{6}=\frac{5}{3}\\x=\frac{-7-17}{6}=-4\end{cases}}\)

a) \(2x-\frac{3x-1}{3}=2+\frac{x-3}{4}\)

<=> 24x - 4(3x - 1) = 24 + 3(x - 3)

<=> 24x - 12x - 4 = 24 + 3x - 9

<=> 12x + 4 = 24 + 3x - 9

<=> 12x + 4 = 3x + 15

<=> 12x = 3x + 15 - 4

<=> 12x = 3x + 11

<=> 12x - 3x = 11

<=> 9x = 11

<=> x = 11/9

Vậy: tập nghiệm phương trình: S = {11/9}

b) \(\frac{x-5}{2}+\frac{1}{4}=\frac{x-2}{3}-x\)

<=> 3(x - 5) + 3/2 = 2(x - 2) - 6x

<=> 3x - 15 + 3/2 = 2x - 4 - 6x

<=> 3x - 27/2 = -4x - 4

<=> 3x = -4x - 4 + 27/2

<=> 3x = -4x + 19/2

<=> 3x + 4x = 19/2

<=> 7x = 19/2

<=> x = 19/14

Vậy: tập nghiệm phương trình: S = {19/14}

c) \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x+2}{8}-5\)

<=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{8}-5\)

<=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{2x+1}{4}-5\)

<=> 2(5x - 3) - 3(7x - 1) = 3(2x + 1) - 60

<=> 10x - 6 - 21x + 3 = 6x + 3 - 60

<=> -11x - 3 = 6x - 57

<=> -3 = 6x - 57 + 11x

<=> -3 = 17x - 57

<=> -3 + 57 = 17x

<=> 54 = 17x

<=> x = 54/17

Vậy: tập nghiệm phương trình: S = {59/17}

d) 3x² + 7x - 20 = 0

<=> 3x² + 12x - 5x - 20 = 0

<=> 3x(x + 4) - 5(x + 4) = 0

<=> (x + 4)(3x - 5) = 0

<=> x + 4 = 0 hoặc 3x - 5 = 0

<=> x = -4 hoặc x = 5/3

Vậy: tập nghiệm phương trình: S = {-4; 5/3}

e) x³ - 3x + 2 = 0

<=> (x² + x - 2)(x - 1) = 0

<=> (x - 1)(x + 2)(x - 1) = 0

<=> x - 1 = 0 hoặc x + 2 = 0

<=> x = 1 hoặc x = -2

Vậy: tập nghiệm phương trình: S = {1; -2}

Giải phương tình nha :v 

a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow2160-198x=160x+12\)

\(\Leftrightarrow358x=2148\)

\(\Leftrightarrow x=6\)

Vậy nghiệm của pt x=6

b)  \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)

\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)

\(\Leftrightarrow-77x-21=48x-396\)

\(\Leftrightarrow125x=375\)

\(\Leftrightarrow3\)

Vậy nghiệm của pt x=3

c)\(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)

\(\Leftrightarrow\frac{15\left(x-3\right)}{20}+\frac{8x-21}{20}=\frac{3x+3}{5}+\frac{30}{5}\)

\(\Leftrightarrow\frac{23x-66}{20}=\frac{3x+33}{5}\)

\(\Leftrightarrow115x-330=60x+660\)

\(\Leftrightarrow55x=990\)

\(\Leftrightarrow x=18\)

Vậy nghiệm của pt x=18

d) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x+3}{4}-\frac{20}{4}=\frac{4\left(3x-1\right)}{10}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x-17}{4}=\frac{9x-6}{10}\)

\(\Leftrightarrow60x-170=36x-24\)

\(\Leftrightarrow24x=146\)

\(\Leftrightarrow x=\frac{73}{12}\)

Vậy nghiệm của pt \(x=\frac{73}{12}\)

ĐK: ` x \ne 2/7`

`(2x+3)((3x+8)/(2-7x)+1)=(x-5)((3x+8)/(2-7x)+1)`

`<=> ((3x+8)(2-7x)+1)(2x+3-x+5)=0`

`<=> ((3x+8)/(2-7x)+1)(x+8)=0`

 \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}=-1\\x+8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

Vậy `S={5/2 ; -8}`.

\(a,4x-6< 7x-12\)

\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)

\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)

\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)

\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)

\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)

\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)

\(\Leftrightarrow-16x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{16}\)

\(d,-12-8x>3+2x-\left(5-7x\right)\)

\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)

\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)

\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)

\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)