Bài 1:
a) ${ \frac{3}{5} + \frac{2}{5}.(\frac{-3}{4})}$
= ${ \frac{3}{5} + \frac{-3}{10}}$
= ${ \frac{6}{10}+\frac{-3}{10}}$
= ${ \frac{3}{10}}$
b) ${\frac{√9}{5}.\frac{1}{2}+\frac{3}{5}.\frac{5}{2}+\frac{3}{5}.\frac{13}{4}}$
= ${\frac{3}{5}.\frac{1}{2}+\frac{3}{5}.\frac{5}{2}+\frac{3}{5}.\frac{13}{4}}$
= ${ \frac{3}{5}.(\frac{1}{2}+\frac{5}{2}+\frac{13}{4})}$
= ${ \frac{3}{5}.(\frac{2}{4}+\frac{10}{4}+\frac{13}{4})}$
= ${ \frac{3}{5}. \frac{25}{4}}$
= ${\frac{15}{4}}$
c) ${ (\frac{-2}{3})^3:\frac{8}{3}-|\frac{-5}{3}|}$
= ${ \frac{-8}{27}:\frac{8}{3}-\frac{5}{3}}$
= ${ \frac{-1}{9} -\frac{5}{3}}$
= ${ \frac{-1}{9}-\frac{15}{9}}$
= ${ \frac{-16}{9}}$
d) ${\frac{8^5.27^3}{6^9.4^4}}$
= ${ \frac{(2^3)^5.(3^3)^3}{3^9.2^9.(2^2)^4}}$
= ${ \frac{2^{15}.3^9}{3^9.2^9.2^8}}$
= ${ \frac{2^{15}.3^9}{3^9.2^{17}}}$
= ${ \frac{1}{2^2}}$
= ${ \frac{1}{4}}$
Bài 2:
a) ${\frac{7}{10}+x=(\frac{-1}{2})^2}$
=> ${\frac{7}{10}+x=\frac{1}{4}}$
=> ${ x = \frac{1}{4}-\frac{7}{10}}$
=> ${ x= \frac{5}{20}-\frac{14}{20}}$
=> ${ x =\frac{-9}{20}}$
b) ${\frac{5}{2}x-\frac{2}{3}=\frac{7}{2}}$
=> ${\frac{5}{2}x = \frac{7}{2}+\frac{2}{3}}$
=> ${\frac{5}{2}x= \frac{25}{6}}$
=> ${ x=\frac{25}{6}:\frac{5}{2}}$
=> ${x=\frac{5}{3}}$
c) ${\frac{4}{5}-|3x-5|=\frac{2}{5}}$
=> ${ |3x-5| = \frac{4}{5}-\frac{2}{5}}$
=> ${ |3x-5| = \frac{2}{5}}$
=> \(\left[ \begin{array}{l}3x-5=\frac{2}{5}\\3x-5=\frac{-2}{5}\end{array} \right.\)
=> \(\left[ \begin{array}{l}3x=\frac{27}{5}\\3x=\frac{23}{5}\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=\frac{9}{5}\\x=\frac{23}{15}\end{array} \right.\)
$\text {Vậy x ∈ }$ {${\frac{9}{5};\frac{23}{15}}$}
