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a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
a, \(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}\right)=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+7\right)}{\left(x-2\right)\left(x+2\right)}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{x^2+3x+2-x^2-5x+14}{x^2-4}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{14-2x}{x^2-4}\right)=0\)
\(\Leftrightarrow12=14-2x\)
\(\Leftrightarrow x=1\)
Vậy x = 1
a.\(x-\dfrac{2}{3}=\dfrac{8}{7}\)
\(x=\dfrac{8}{7}+\dfrac{2}{3}\)
x=\(\dfrac{38}{21}\)
b.\(\left(x+\dfrac{1}{3}\right)=\dfrac{4}{25}
\)
x=\(\dfrac{4}{25}-\dfrac{1}{3}\)
x=\(-\dfrac{13}{75}\)
c.\(-\dfrac{2}{3}:x+\dfrac{5}{8}=-\dfrac{7}{12}\)
\(-\dfrac{2}{3}:x=-\dfrac{29}{24}\)
x=\(\dfrac{16}{29}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)
d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)
a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)
b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)
c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)
\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)
Vậy....
b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)
\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)
\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)
Vậy....
c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)
\(\dfrac{x}{182}=-\dfrac{5}{26}\)
\(=>x\cdot26=-5\cdot182\)
\(26x=-910\)
\(x=-910:26=-35\)
Vậy....
a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)
\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)
\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)
hay \(x=\dfrac{37}{70}\)
Vậy: \(x=\dfrac{37}{70}\)
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
a) \(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{2}{5}\\\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{22}{15}\\x=-\dfrac{2}{15}\end{matrix}\right.\)
b) \(\Rightarrow\left(1-\dfrac{1}{4}x\right)^2=\dfrac{121}{49}\)
\(\Rightarrow\left[{}\begin{matrix}1-\dfrac{1}{4}x=\dfrac{11}{7}\\1-\dfrac{1}{4}x=-\dfrac{11}{7}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{16}{7}\\x=\dfrac{72}{7}\end{matrix}\right.\)