#andy

\[\begin{array}{l}
e)\dfrac{{3\left( {x - 11} \right)}}{4} = \dfrac{{3\left( {x + 1} \right)}}{5} - \dfrac{{2\left( {2x - 5} \right)}}{{10}}\\
MC:20\\
 \Leftrightarrow \dfrac{{15\left( {x - 11} \right)}}{{20}} = \dfrac{{12\left( {x + 1} \right)}}{{20}} - \dfrac{{4\left( {2x - 5} \right)}}{{20}}\\
 \Leftrightarrow \dfrac{{15x - 165}}{{20}} = \dfrac{{12x + 12}}{{20}} - \dfrac{{8x - 20}}{{20}}\\
 \Leftrightarrow 15x - 165 = 12x + 12 - 8x + 20\\
 \Leftrightarrow 15x - 12x + 8x = 12 + 20 + 165\\
 \Leftrightarrow 11x = 197\;\\
 \Leftrightarrow x = \dfrac{{197}}{{11}}\\
 \Rightarrow S = \left\{ {\dfrac{{197}}{{11}}} \right\}\\
h)\dfrac{{2x - 5}}{6} - x + 2 = \dfrac{{5x - 3}}{3} - \dfrac{{6x - 7}}{4} + x\\
 \Leftrightarrow \dfrac{{2\left( {2x - 5} \right)}}{{12}} - \dfrac{{12x}}{{12}} + \dfrac{{24}}{{12}} = \dfrac{{4\left( {5x - 3} \right)}}{{12}} - \dfrac{{3\left( {6x - 7} \right)}}{{12}} + \dfrac{{12x}}{{12}}\\
 \Leftrightarrow \dfrac{{4x - 10}}{{12}} - \dfrac{{12x}}{{12}} + \dfrac{{24}}{{12}} = \dfrac{{20x - 12}}{{12}} - \dfrac{{18x - 21}}{{12}} + \frac{{12x}}{{12}}\\
 \Leftrightarrow 4x - 10 - 12x + 24 = 20x - 12 - 18x + 21 + 12x\\
 \Leftrightarrow 4x - 12x - 20x + 18x - 12x =  - 12 + 21 + 10 - 24\\
 \Leftrightarrow  - 22x =  - 5\\
 \Leftrightarrow x = \dfrac{5}{{22}}\\
 \Rightarrow S = \left\{ {\dfrac{5}{{22}}} \right\}
\end{array}\]

Đáp án + Giải thích các bước giải:

`e)` 

`(3(x-11))/4=(3(x+1))/5-(2(2x-5))/10 `

`⇔(15(x-11))/20=(12(x+1))/20-(4(2x-5))/20 `

`⇔(15x-165)/20=(12x+12)/20-(8x-20)/20 `

`⇔15x-165=12x+12-8x+20 `

`⇔15x-12x+8x=12+20+165 `

`⇔11x=197  ⇔x=197/11 `

Vậy `S={197/11} `

`h)` 

`(2x-5)/6-x+2=(5x-3)/3-(6x-7)/4+x `

`⇔(2(2x-5))/12-(12x)/12+24/12=(4(5x-3))/12-(3(6x-7))/12+(12x)/12 `

`⇔(4x-10)/12-(12x)/12+24/12=(20x-12)/12-(18x-21)/12+(12x)/12 `

`⇔4x-10-12x+24=20x-12-18x+21+12x `

`⇔4x-12x-20x+18x-12x=-12+21+10-24 `

`⇔-22x=-5 `

`⇔x=5/22 `

Vậy `S={5/22}`