4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

c: \(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}=\dfrac{-2}{x^2}\)

`@` `\text {Ans}`

`\downarrow`

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

`=> (x-3)5 = (2x+1)3`

`=> 5x-15 = 6x+3`

`=> 5x-6x = 15+3`

`=> -x=18`

`=> x=-18`

\(\dfrac{x+1}{22}=\dfrac{6}{x}\)

`=> (x+1)x = 22*6`

`=> (x+1)x = 132`

`=> x^2 + x = 132`

`=> x^2+x-132=0`

`=> (x-11)(x+12)=0`

`=>`\(\left[{}\begin{matrix}x-11=0\\x+12=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=11\\x=-12\end{matrix}\right.\)

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

`=> (2x-1)x = 2*5`

`=> 2x^2 - x =10`

`=> 2x^2 - x - 10 =0`

`=> 2x^2 + 4x - 5x - 10 =0`

`=> (2x^2 + 4x) - (5x+10)=0`

`=> 2x(x+2) - 5(x+2)=0`

`=> (2x-5)(x+2)=0`

`=>`\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)

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

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\)

`=> (2x-1)(2x+1)=21*3`

`=> 4x^2 + 2x - 2x - 1 = 63`

`=> 4x^2 - 1=63`

`=> 4x^2 - 1 - 63=0`

`=> 4x^2 - 64 = 0`

`=> 4(x^2 - 16)=0`

`=> 4(x^2 + 4x - 4x - 16)=0`

`=> 4[(x^2+4x)-(4x+16)]=0`

`=> 4[x(x+4)-4(x+4)]=0`

`=> 4(x-4)(x+4)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

\(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\)

`=> (2x+1)(x+1) = 9*5`

`=> (2x+1)(x+1)=45`

`=> 2x^2 + 2x + x + 1 = 45`

`=> 2x^2 + 3x + 1 =45`

`=> 2x^2 + 3x + 1 - 45 =0`

`=> 2x^2+3x-44=0`

`=> 2x^2 + 11x - 8x - 44=0`

`=> (2x^2 +11x) - (8x+44)=0`

`=> x(2x+11) - 4(2x+11)=0`

`=> (x-4)(2x+11)=0`

`=>`\(\left[{}\begin{matrix}x-4=0\\2x+11=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

\(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \left(x-3\right)\cdot5=\left(2x+1\right)\cdot3\\ x5-15=6x+3\\ x5-6x=3+15\\ -x=18\\ \Rightarrow x=-18\)

\(\dfrac{x+1}{22}=\dfrac{6}{x}\\ \left(x+1\right)\cdot x=6\cdot22\\ \left(x+1\right)\cdot x=2\cdot3\cdot2\cdot11\\ \left(x+1\right)\cdot x=12\cdot11\\ \Rightarrow x=11\)

\(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\\ \left(2x-1\right)\cdot\left(2x+1\right)=21\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot3\cdot3\\ \left(2x-1\right)\cdot\left(2x+1\right)=7\cdot9\\ \Rightarrow2x+1=9\\ 2x=8\\ x=4\)

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

\(\Leftrightarrow2x-8+12x=4x-2\)

\(\Leftrightarrow10x=6\)

hay \(x=\dfrac{3}{5}\)

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

\(\Leftrightarrow15x-6-30=10-20x\)

\(\Leftrightarrow35x=46\)

hay \(x=\dfrac{46}{35}\)

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

\(\Leftrightarrow3x-6-4=6x-6\)

\(\Leftrightarrow-3x=4\)

hay \(x=-\dfrac{4}{3}\)

1)\(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)

\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)

\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)

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

\(\Leftrightarrow40x-20+45x-30=48x-36\)

\(\Leftrightarrow37x=14\)

hay \(x=\dfrac{14}{37}\)

5: Ta có: \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)

\(\Leftrightarrow2x-6-3x-6=x+4-9\)

\(\Leftrightarrow-x-x=-5-12=-17\)

hay \(x=\dfrac{17}{2}\)

Mk giải giúp bạn phần a thôi nha! (Dài lắm, lười :v)

a, 1 + \(\dfrac{x}{3-x}\) = \(\dfrac{5x}{\left(x+2\right)\left(x+3\right)}+\dfrac{2}{x+2}\) (x \(\ne\) -2; x \(\ne\) \(\pm\) 3)

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

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

\(\Leftrightarrow\) \(\dfrac{3}{3-x}=\dfrac{7x+6}{x^2+5x+6}\)

Vì 3 - x \(\ne\) 0; x² + 5x + 6 \(\ne\) 0

\(\Rightarrow\) 3(x² + 5x + 6) = (7x + 6)(3 - x)

\(\Leftrightarrow\) 3x² + 15x + 18 = 21x - 7x² + 18 - 6x

\(\Leftrightarrow\) 10x² = 0

\(\Leftrightarrow\) x = 0 (TM)

Vậy S = {0}

Chúc bn học tốt! (Nếu bạn cần phần nào khác mk có thể giúp bn chứ đừng có đăng hết lên, ít người làm lắm :v)

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

vậy..

d)\(\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(2-x\right)}+\dfrac{\left(x+3\right)\left(x-3\right)}{\left(2-x\right)\left(x-3\right)}=0\\ \Leftrightarrow5+x^2-9=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

vậy..

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)

\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)

\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

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

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1