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\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2\left(2x+1\right)}-\frac{2.3}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(2x+1=13\)
\(\Rightarrow x=6\)
\(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\frac{3}{2x+1}+\frac{2.5}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\) - \(\frac{6}{6x+3}\) = \(\frac{12}{26}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{2.5}{2\left(2x+1\right)}\) - \(\dfrac{2.3}{3\left(2x+1\right)}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) - \(\dfrac{2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{3}{2x+1}\) + \(\dfrac{5}{2x+1}\) + \(\dfrac{-2}{2x+1}\) = \(\dfrac{6}{13}\)
\(\dfrac{6}{2x+1}\) = \(\dfrac{6}{13}\)
\(\Rightarrow\) (2x+1).6 = 6.13 2x + 1 = \(\dfrac{6.13}{6}\) 2x + 1 = 13 2x = 13 - 1 2x = 12 x = 12:2 x = 6 Vậy x = 6
Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}\)= \(\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}\)
= \(\dfrac{3+5-2}{2x+1}=\dfrac{6}{2x+1}=\dfrac{12}{26}\) \(\Rightarrow156=24x+12\Rightarrow24x=144\Rightarrow x=6\)
Vậy x=6.
Học tốt nha 
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow70=-6\left(x-1\right)\)
\(\Rightarrow6x=6-70\)
\(\Rightarrow6x=-64\)
\(\Rightarrow x=\frac{-32}{3}x\ne1\)
a) \(\frac{2}{3}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\)
\(\frac{x}{18}\le\frac{7}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\)
tu tim x o 2 truong hop tren
b) de \(\frac{11}{2x+1}\) nguyen thi \(2x+1\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
2x+1=-1 suy ra x=-1
2x+1=1 suy ra x=0
2x+1=11 suy ra x=5
2x+1=-11 suy ra x=-6
Vay de ......thi x thuoc {-1;0;5;6}
Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)
\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)
=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)
=> \(\frac{6}{2x+1}=\frac{6}{13}\)
=> 2x + 1 = 13
=> 2x = 12
=> x = 6
Vậy x = 6
\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)
\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)
\(\frac{6}{2x+1}=\frac{6}{13}\)
\(\Rightarrow2x+1=13\left(6=6\right)\)
\(2x=12\)
\(x=6\)