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\(\left|5x+13\right|=2x-7\)
khi \(x>\frac{7}{2}\), biểu thức có dạng:
\(\orbr{\begin{cases}5x+13=2x-7\\5x+13=7-2x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=-20\\7x=-6\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{20}{3}\\x=-\frac{6}{7}\end{cases}}}\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
\(\frac{-1}{10}\)-\(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{1}{10}\)
\(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-1}{10}\)-\(\frac{1}{10}\)
\(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-2}{10}\)
\(\frac{2}{5}\)x =\(\frac{-4}{20}\)-\(\frac{7}{20}\)
\(\frac{2}{5}\)x =\(\frac{-11}{20}\)
x =\(\frac{-11}{20}\).\(\frac{5}{2}\)
x =-1\(\frac{3}{8}\)
vậy...
-1/10-2/5x+7/20=1/10
2/5x+7/10=-1/10-1/10
2/5x+7/10=0
2/5x=0-7/10
2/5x=-3/10
x=-3/10:2/5
x=-3/4