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a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
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
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
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: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)
\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)
\(=\dfrac{4}{6}=\dfrac{2}{3}\)
b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)
\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)
\(=\dfrac{31}{24}\)
c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)
d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)
\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)
a) Ta có: \(\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{20}\right)\)
\(=\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)\cdot\left(\dfrac{5}{20}-\dfrac{4}{20}-\dfrac{1}{20}\right)\)
\(=0\cdot\left(\dfrac{617}{191}+\dfrac{29}{33}-\dfrac{115}{17}\right)=0\)
b) Ta có: \(\dfrac{12}{5}\cdot\left(\dfrac{10}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\left(\dfrac{40}{12}-\dfrac{5}{12}\right)\)
\(=\dfrac{12}{5}\cdot\dfrac{35}{12}\)
=7