giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

\(\dfrac{3}{5}x+\dfrac{1}{2}=\dfrac{1}{5}x+\dfrac{5}{6}\\ \Rightarrow\dfrac{3}{5}x+\dfrac{1}{2}-\dfrac{1}{5}x-\dfrac{5}{6}=0\\ \Rightarrow\left(\dfrac{3}{5}x-\dfrac{1}{5}x\right)+\left(\dfrac{1}{2}-\dfrac{5}{6}\right)=0\\ \Rightarrow\dfrac{2}{5}x+\dfrac{3}{6}-\dfrac{5}{6}=0\\ \Rightarrow\dfrac{2}{5}x-\dfrac{2}{6}=0\\ \Rightarrow\dfrac{2}{5}x=\dfrac{2}{6}\\ \Rightarrow x=\dfrac{2}{6}:\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{6}.\dfrac{5}{2}\\ \Rightarrow x=\dfrac{5}{6}\)

a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27

b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4

tick cho mik nhé 

a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3

b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x² => 4² = x² => x=4

        Tick cho mình đi 

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

`@` `\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`

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

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

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

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

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

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

Lời giải:

a) 

$\frac{4}{7}x=\frac{2}{3}+\frac{1}{5}=\frac{13}{15}$

$x=\frac{13}{15}:\frac{4}{7}=\frac{91}{60}$

b) 

$\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}$

$\frac{5}{7}:x=\frac{-19}{30}$

$x=\frac{5}{7}:\frac{-19}{30}=\frac{-150}{133}$

a) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\) 

            \(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\) 

            \(\dfrac{4}{7}.x=\dfrac{13}{15}\) 

                \(x=\dfrac{13}{15}:\dfrac{4}{7}\) 

                \(x=\dfrac{91}{60}\) 

b) \(\dfrac{4}{5}+\dfrac{5}{7}:x=\dfrac{1}{6}\) 

            \(\dfrac{5}{7}:x=\dfrac{1}{6}-\dfrac{4}{5}\) 

            \(\dfrac{5}{7}:x=\dfrac{-19}{30}\) 

                  \(x=\dfrac{5}{7}:\dfrac{-19}{30}\) 

                 \(x=\dfrac{-150}{133}\)

a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)

b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)

c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)

d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)

\(a,\)\(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(x=\dfrac{13}{7}.\dfrac{6}{13}\)

\(x=\dfrac{6}{7}\)

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

\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)

\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)

\(\dfrac{4}{7}.x=\dfrac{13}{15}\)

\(x=\dfrac{13}{15}:\dfrac{4}{7}\)

\(x=\dfrac{13}{15}.\dfrac{7}{4}\)

\(x=\dfrac{91}{60}\)

a: Ta có: \(x:\dfrac{6}{13}=\dfrac{13}{7}\)

\(\Leftrightarrow x=\dfrac{13}{7}\cdot\dfrac{6}{13}\)

hay \(x=\dfrac{6}{7}\)

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

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{13}{15}\)

hay \(x=\dfrac{91}{60}\)

a) \(x=0,05\)

b) \(x=1,125\)

c) \(x=0,96\)

d) \(x=0,025\)

Bạn tự làm đi dễ mà . Cố mag vận động đầu óc đừng copy làm bài nữa khó lắm mới hỏi thôi

Chuẩn đấy

\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)