a, \(\frac{-5}{2}\) x \(\frac{-14}{7}\) = \(\frac{-5}{2}\) x -2 = 5
b , \(\left(1-\frac{1}{4}\right)^2+\left|\frac{-5}{4}\right|+\frac{-7}{8}-\frac{9}{16}\)
= \(\left(\frac{3}{4}\right)^2+\frac{5}{4}+\frac{-7}{8}-\frac{9}{16}\)
= \(\frac{9}{16}+\frac{5}{4}+\frac{-7}{8}-\frac{9}{16}\)
= \(\left(\frac{9}{16}-\frac{9}{16}\right)+\left(\frac{10}{8}+\frac{-7}{8}\right)\)
= \(\frac{3}{8}\)
BT2
a, \(x+\frac{2}{5}=\frac{-4}{3}\)
\(x=\frac{-4}{3}-\frac{2}{5}\)
\(x=\frac{-25}{15}\)
b, \(\left|\frac{2}{3}-x\right|=\frac{6}{5}\)
=> \(\frac{2}{3}-x=\frac{6}{5}\) hoặc \(\frac{2}{3}-x=\frac{-6}{5}\)
x = \(\frac{2}{3}-\frac{6}{5}\) x = \(\frac{2}{3}-\frac{-6}{5}\)
\(x=\frac{-8}{15}\) \(x=\frac{28}{15}\)
vậy x = \(\frac{-8}{15}\) hoặc x = \(\frac{28}{15}\)
