a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)

c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.

a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)

b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)

a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)

\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)

\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)

b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)

=459/940

a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)

b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)

c) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)

\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)

\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)

\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)

=1

Trời ơi cái đề bài !!!

Thoy thì làm từng câu vậy

a) \(I=10101.\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{111111}\right)\)

\(I=10101.\left(\dfrac{10}{222222}+\dfrac{5}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\left(\dfrac{15}{222222}-\dfrac{8}{222222}\right)\)

\(I=10101.\dfrac{7}{222222}\)

\(I=\dfrac{7}{22}\)

1.

a) \(A=10101\left(\dfrac{5}{111111}+\dfrac{5}{222222}-\dfrac{4}{3.7.11.13.37}\right)\)\(A=\dfrac{10101.5}{10101.11}+\dfrac{10101.5}{10101.22}-\dfrac{10101.4}{10101.11}\)

\(A=\dfrac{5}{11}+\dfrac{5}{22}-\dfrac{4}{11}=\dfrac{7}{22}\)

1. Tính hợp lí

a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)

\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)

\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)

\(=1+\dfrac{-7}{19}\)

\(=\dfrac{12}{19}\)

b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)

\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)

\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)

\(=2.\dfrac{-5}{2}\)

\(=-5\)

c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)

\(=\dfrac{3}{5}.-1\)

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

d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)

\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)

\(=\dfrac{7}{4}.1\)

\(=\dfrac{7}{4}\)

Chúc bạn học tốt

a) $0,7+\genfrac{}{}{0ex}{}{-7}{19}-\left(-0,3\right)$

$=\genfrac{}{}{0ex}{}{7}{10}+\genfrac{}{}{0ex}{}{-7}{19}+\genfrac{}{}{0ex}{}{3}{10}$

$=\left(\genfrac{}{}{0ex}{}{7}{10}+\genfrac{}{}{0ex}{}{3}{10}\right)+\genfrac{}{}{0ex}{}{-7}{19}$

$=1+\genfrac{}{}{0ex}{}{-7}{19}$

$=\genfrac{}{}{0ex}{}{12}{19}$

b) $\genfrac{}{}{0ex}{}{5}{3}.\left(-2,5\right):\genfrac{}{}{0ex}{}{5}{6}$

$=\genfrac{}{}{0ex}{}{5}{3}.\genfrac{}{}{0ex}{}{-5}{2}.\genfrac{}{}{0ex}{}{6}{5}$

$=\left(\genfrac{}{}{0ex}{}{5}{3}.\genfrac{}{}{0ex}{}{6}{5}\right).\genfrac{}{}{0ex}{}{-5}{2}$

$=2.\genfrac{}{}{0ex}{}{-5}{2}$

$=-5$

c) $0,6.\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{12}{17}$

$=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{12}{17}$

$=\genfrac{}{}{0ex}{}{3}{5}.\left(\genfrac{}{}{0ex}{}{-5}{17}-\genfrac{}{}{0ex}{}{12}{17}\right)$

$=\genfrac{}{}{0ex}{}{3}{5}.-1$

$=\genfrac{}{}{0ex}{}{-3}{5}$

d) $\genfrac{}{}{0ex}{}{7}{4}.\genfrac{}{}{0ex}{}{5}{2}-\genfrac{}{}{0ex}{}{7}{4}.\genfrac{}{}{0ex}{}{3}{2}$

$=\genfrac{}{}{0ex}{}{7}{4}.\left(\genfrac{}{}{0ex}{}{5}{2}-\genfrac{}{}{0ex}{}{3}{2}\right)$

$=\genfrac{}{}{0ex}{}{7}{4}.1$

$=\genfrac{}{}{0ex}{}{7}{4}$

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