a: \(-\frac32-2x+\frac34=-2\)
=>\(-2x-\frac64+\frac34=-2\)
=>\(-2x-\frac34=-2\)
=>\(2x+\frac34=2\)
=>\(2x=2-\frac34=\frac54\)
=>\(x=\frac58\)
b: \(\left(-\frac23x-\frac35\right)\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac25\)
=>\(\left(-\frac23x-\frac35\right)\left(-\frac32-\frac{10}{3}\right)=\frac25\)
=>\(\left(\frac23x+\frac35\right)\left(\frac32+\frac{10}{3}\right)=\frac25\)
=>\(\left(\frac23x+\frac35\right)=\frac25:\frac{29}{6}=\frac25\cdot\frac{6}{29}=\frac{12}{145}\)
=>\(\frac23x=\frac{12}{145}-\frac35=\frac{12}{145}-\frac{87}{145}=\frac{-75}{145}=\frac{-15}{29}\)
=>\(x=-\frac{15}{29}:\frac23=-\frac{15}{29}\cdot\frac32=\frac{-45}{58}\)
c: \(\frac{x}{2}-\left(\frac35x-\frac{13}{5}\right)=-\left(\frac75+\frac{7}{10}x\right)\)
=>\(\frac{x}{2}-\frac35x+\frac{13}{5}=-\frac75-\frac{7}{10}x\)
=>\(\frac{5x}{10}-\frac{6x}{10}+\frac{13}{5}=-\frac75-\frac{7}{10}x\)
=>\(\frac{6}{10}x=-\frac75-\frac{13}{5}=-\frac{20}{5}=-4\)
=>\(x=-4:\frac35=-4\cdot\frac53=-\frac{20}{3}\)
d: \(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac13\)
=>\(\frac{2\left(2x-3\right)-9}{6}=\frac{5-3x-2}{6}=\frac{-3x+3}{6}\)
=>2(2x-3)-9=-3x+3
=>4x-6-9=-3x+3
=>7x=3+6+9=18
=>\(x=\frac{18}{7}\)
e: \(\frac{2}{3x}-\frac{3}{12}=\frac45-\left(\frac{7}{x}-2\right)\) (ĐKXĐ: x<>0)
=>\(\frac{2}{3x}=\frac45+\frac14-\frac{7}{x}+2=\frac{16+5+40}{20}-\frac{7}{x}=\frac{61}{20}-\frac{7}{x}\)
=>\(\frac{2}{3x}+\frac{21}{3x}=\frac{61}{20}\)
=>\(\frac{63}{3x}=\frac{61}{20}\)
=>\(\frac{21}{x}=\frac{61}{20}\)
=>\(x=21\cdot\frac{20}{61}=\frac{420}{61}\) (nhận)
f: ĐKXĐ: x<>1
\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac34-\frac65\right)=\frac{5}{2-2x}\)
=>\(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{15}{20}-\frac{24}{20}\right)=\frac{-5}{2\left(x-1\right)}\)
=>\(\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac23\left(\frac{15}{20}-\frac{24}{20}\right)=\frac23\cdot\frac{-9}{20}=\frac{-18}{60}=-\frac{3}{10}\)
=>\(\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
=>2(x-1)=-70/3
=>\(x-1=-\frac{35}{3}\)
=>\(x=-\frac{32}{3}\) (nhận)
g: ĐKXĐ: x<>3/2
\(3-\frac{2}{2x-3}=\frac25+\frac{2}{9-6x}-\frac32\)
=>\(3-\frac{2}{2x-3}=\frac{4}{10}-\frac{15}{10}-\frac{2}{6x-9}\)
=>\(3-\frac{2}{2x-3}=\frac{-11}{10}-\frac{2}{3\left(2x-3\right)}\)
=>\(-\frac{2}{2x-3}+\frac{2}{3\left(2x-3\right)}=\frac{-11}{10}-3\)
=>\(\frac{-4}{3\left(2x-3\right)}=\frac{-11-30}{10}=\frac{-41}{10}\)
=>\(3\left(2x-3\right)=\frac{40}{41}\)
=>\(2x-3=\frac{40}{123}\)
=>\(2x=\frac{40}{123}+3=\frac{40+3\cdot123}{123}=\frac{409}{123}\)
=>\(x=\frac{409}{246}\) (nhận)
h: ĐKXĐ: x<>0
\(\frac{x}{2}-\frac{1}{x}=\frac{1}{12}\)
=>\(\frac{x^2-2}{2x}=\frac{1}{12}\)
=>\(6\left(x^2-2\right)=x\)
=>\(6x^2-x-12=0\)
=>\(6x^2-9x+8x-12=0\)
=>3x(2x-3)+4(2x-3)=0
=>(2x-3)(3x+4)=0
=>x=3/2(nhận) hoặc x=-4/3(nhận)
i: \(x^2-\frac76x+\frac13=0\)
=>\(6x^2-7x+2=0\)
=>\(6x^2-4x-3x+2=0\)
=>2x(3x-2)-(3x-2)=0
=>(3x-2)(2x-1)=0
=>\(\left[\begin{array}{l}3x-2=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=2\\ 2x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac23\\ x=\frac12\end{array}\right.\)
m: \(\left(\frac32-\frac{2}{-5}\right):x-\frac12=\frac32\)
=>\(\left(\frac32+\frac25\right):x=\frac32+\frac12=2\)
=>\(\left(\frac{15}{10}+\frac{4}{10}\right):x=2\)
=>\(\frac{19}{10}:x=2\)
=>\(x=\frac{19}{10}:2=\frac{19}{20}\)
n: \(\left(\frac32-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(-\frac34+\frac{5}{22}+\frac{3}{26}\right)\)
=>\(\left(\frac32-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=-\frac12\left(\frac32-\frac{5}{11}-\frac{3}{13}\right)\)
=>\(2x-2=-\frac12\)
=>\(2x=2-\frac12=\frac32\)
=>\(x=\frac34\)
