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28 tháng 3 2019

\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)

\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)

\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)

\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)

\(\Leftrightarrow B=\frac{31}{20}\)

28 tháng 3 2019

Ai tui sai là người ngu nhất thế giới.22

28 tháng 4 2018

a)\(\frac{-10}{13}+\frac{8}{17}-\frac{3}{13}+\frac{12}{17}-\frac{11}{20}\)                

\(\frac{-10}{13}+\frac{8}{17}+\frac{-3}{13}+\frac{12}{17}+\frac{-11}{20}\)     

=\(\left(\frac{-10}{13}+\frac{-3}{13}\right)+\left(\frac{8}{17}+\frac{12}{17}\right)+\frac{-11}{20}\)

=\(\frac{-13}{13}+\frac{20}{17}+\frac{-11}{20}\)

\(\frac{-127}{340}\)

b) \(\frac{3}{4}+\frac{-5}{6}-\frac{11}{-12}\)

\(\frac{3}{4}+\frac{-5}{6}+\frac{11}{12}\)

\(\frac{9}{12}+\frac{-10}{12}+\frac{11}{12}\)

=\(\frac{10}{12}=\frac{5}{6}\)

c) \(\left[13.\frac{4}{9}+2.\frac{1}{9}\right]-3.\frac{4}{9}\)

\(13+2.\left(\frac{4}{9}+\frac{1}{9}\right)-3.\frac{4}{9}\)

=\(15.\frac{5}{9}-3.\frac{4}{9}\)

=\(\left[15-3.\left(\frac{5}{9}-\frac{4}{9}\right)\right]\)

=\(12.\frac{1}{9}\)

=\(\frac{4}{3}\)

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Bài 1:

a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)

\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)

\(=\frac{-3}{5}\)

b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)

\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)

\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)

c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)

\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)

\(=3+2=5\)

d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)

\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)

\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)

e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)

\(=-1+1+\frac{-7}{9}\)

\(=-\frac{7}{9}\)

f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)

\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)

\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)

26 tháng 7 2016

p) \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)

\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)

\(=-1+1+\frac{-2}{11}\)

\(=-\frac{2}{11}\)

q) \(\frac{5}{13}+\frac{-5}{17}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)

\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{17}\)


\(=1+\left(-1\right)+\frac{-5}{17}=\frac{-5}{17}\)

r) \(\frac{1}{5}+\frac{-2}{9}+\frac{-7}{9}+\frac{4}{5}+\frac{16}{17}\)

\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+\frac{-7}{9}\right)+\frac{16}{17}\)

\(=1+\left(-1\right)+\frac{16}{17}=\frac{16}{17}\)

8 tháng 2 2019

Đặt \(A=\frac{9+\frac{9}{11}+\frac{18}{23}-\frac{27}{37}}{8+\frac{8}{11}+\frac{16}{23}-\frac{24}{37}}-\frac{2+\frac{16}{29}-\frac{24}{13}-\frac{32}{11}}{3+\frac{24}{29}-\frac{36}{13}-\frac{48}{11}}\)\(=\frac{9\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}{8\left(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37}\right)}-\frac{2\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}{3\left(1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\right)}\)

\(=\frac{9}{8}-\frac{2}{3}\)(do \(1+\frac{1}{11}+\frac{2}{23}-\frac{3}{37};1+\frac{8}{29}-\frac{12}{13}-\frac{16}{11}\ne0\))

\(=\frac{27}{24}-\frac{16}{24}=\frac{11}{24}.\)

Vậy A = \(\frac{11}{24}.\)

31 tháng 8 2017

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{1454}{323}+\frac{35}{43}+6\)

\(=5,...+6\)

\(=11,...\)

3 tháng 7 2018

\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)

\(=\sqrt{3}-2\) 

\(VayA=\sqrt{3}-2\)