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20 tháng 1 2022

\(\left(-\frac{1}{3}\right)^2-\left(-\frac{12}{17}\right)^0+\left(-\frac{1}{2}\right)^3.4^2\)

\(=\)\(\frac{1}{9}-1+\frac{-1}{8}.16\)

\(=\)\(\frac{1}{9}-1+\frac{-16}{8}\)

\(=\)\(\frac{1}{9}-1+-2\)

\(=\)\(\frac{-8}{9}+-2\)

\(=\)\(\frac{-24}{9}\)

17 tháng 12 2016

a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)

= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)

= \(\frac{17}{9}-\frac{2}{3}\)

= \(\frac{11}{9}\)

b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)

= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)

= \(\frac{2}{5}.\frac{7}{12}\)

= \(\frac{7}{30}\)

Mình lười làm quá, hay mình nói kết quả cho bn thôi nha

c) -6

d) 3

e) 3

g) 12

h) \(\frac{23}{18}\)

i) \(\frac{-69}{20}\)

k) \(\frac{-1}{2}\)

l) \(\frac{49}{5}\)

4 tháng 10 2021

yutyugubhujyikiu

a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)

\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)

b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)

\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)

17 tháng 1 2020

b) \(\frac{\frac{2}{3}+\frac{5}{7}+\frac{4}{21}}{\frac{5}{6}+\frac{11}{7}-\frac{7}{21}}\)

\(=\frac{\frac{29}{21}+\frac{4}{21}}{\frac{101}{42}-\frac{7}{21}}\)

\(=\frac{\frac{11}{7}}{\frac{29}{14}}\)

\(=\frac{22}{29}.\)

Chúc bạn học tốt!

4 tháng 7 2019

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

\(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\) 

\(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)

\(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)

\(\frac{15}{8}+\frac{5}{8}\)

\(\frac{5}{2}\)

4 tháng 7 2019

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)

\(=\frac{49}{34}\)