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\(\frac{x-1}{21}=\frac{3}{x+1}\)
=> \(\left(x-1\right)\left(x+1\right)=21\cdot3\)
=> \(x^2-1=63\)
=> \(x^2=64\)
=> \(\orbr{\begin{cases}x^2=8^2\\x^2=\left(-8\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
=> \(\frac{12}{13}x=2\)
=> \(x=\frac{13}{6}\)
d, \(\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)
\(\Leftrightarrow x^2-1=63\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)
e, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)
-5.(x+1/5) -1/2.(x-2/3)=3/2x-5/6
-5x + (-1) -1/2x -1/3=3/2x-5/6
-5x-1/2x-3/2x=1+1/3-5/6
x.(-5-1/2-3/2)= 6/6+2/6+(-5/6)
x.(-10/2+(-1/2)+(-3/2))=3/6
x.6/2=1/2
x=1/2:6/2
x=1/6
Vậy x = 1/6
a, \(\frac{x}{3}-\frac{1}{4}=-\frac{5}{6}\Leftrightarrow\frac{x}{3}+\frac{7}{12}=0\Leftrightarrow\frac{4x}{12}+\frac{7}{12}=0\)
Khử mẫu ta đc : \(4x+7=0\Leftrightarrow4x=-7\Leftrightarrow x=-\frac{7}{4}\)
b, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow\frac{x+3}{15}=\frac{5}{15}\)
Khử mẫu ta đc : \(x+3=5\Leftrightarrow x=2\)
a, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\Leftrightarrow\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)
b, \(\frac{x-12}{4}=\frac{9-3x}{x}\)
\(\Leftrightarrow x^2-12x=36-12x\Leftrightarrow x^2-12x-36+12x=0\)
\(\Leftrightarrow x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)
c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)
\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)
\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)
\(2+\frac{3}{4}x=\frac{21}{8}\)
\(\frac{3}{4}x=\frac{21}{8}-2\)
\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)
\(\frac{3}{4}x=\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{3}{4}\)
\(x=\frac{5}{8}.\frac{4}{3}\)
\(x=\frac{5}{6}\)
Vậy \(x=\frac{5}{6}\).
d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)
\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)
\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)
\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)
Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).
\(\frac{-5}{6}-x=\frac{7}{12}+\frac{-1}{3}\)
\(\frac{-5}{6}-x=\frac{7}{12}-\frac{4}{12}\)
\(\frac{-5}{6}-x=\frac{1}{4}\)
\(x=\frac{-5}{6}-\frac{1}{4}\)
\(x=\frac{-20}{24}-\frac{6}{24}\)
x=\(\frac{-13}{12}\)
Vậy x=\(\frac{-13}{12}\)
chúc bạn học tốt~