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`2/(3.5) + 2/(5.7) + 2/(7.9) + .... + 2/(37.39)`
`=2 . (1/(3.5) + 1/(5.7) + .... + 1/(37.39)`
`=2 .[ 1/2 .(1/3 -1/5 + 1/5 -1/7 +....+1/37 -1/39)]`
`= 1/3 - 1/39`
`=4/13`
Câu 1: \(-7/6+15/6=8/6=4/3\)
Câu 2: \(x=7/12-1/6=>x=7/12-2/12=>x=5/12\) nhé =)
câu 1
\(\dfrac{-7}{6}+\dfrac{15}{6}=\dfrac{-7+15}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)
câu 2
\(x+\dfrac{1}{6}=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}-\dfrac{1}{6}\)
\(x=\dfrac{7}{12}-\dfrac{2}{12}=\dfrac{7-2}{12}\)
\(x=\dfrac{5}{12}\)
\(-\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{-3}{5}\times\dfrac{4}{7}+\dfrac{-3}{5}\times\dfrac{6}{7}\)
\(=-\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}+\dfrac{6}{7}\right)\)
\(=-\dfrac{3}{5}\times\dfrac{15}{7}=-\dfrac{9}{7}\)
c) Ta có: \(\dfrac{5}{9}\cdot\dfrac{7}{13}-\dfrac{30}{23}+\dfrac{6}{9}\cdot\dfrac{5}{13}+\dfrac{7}{23}\)
\(=\dfrac{5}{13}\cdot\dfrac{7}{9}+\dfrac{6}{9}\cdot\dfrac{5}{13}-\dfrac{30}{23}+\dfrac{7}{23}\)
\(=\dfrac{5}{13}\left(\dfrac{7}{9}+\dfrac{6}{9}\right)-1\)
\(=\dfrac{5}{13}\cdot\dfrac{13}{9}-1\)
\(=\dfrac{-4}{9}\)
d) Ta có: \(\left(-3.2\right)\cdot\dfrac{-15}{64}+\left(0.8-2\dfrac{4}{5}\right):3\dfrac{2}{3}\)
\(=\dfrac{16}{5}\cdot\dfrac{15}{64}+\left(\dfrac{4}{5}-\dfrac{14}{5}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}-2:\dfrac{11}{3}\)
\(=\dfrac{3}{4}-2\cdot\dfrac{3}{11}\)
\(=\dfrac{3}{4}-\dfrac{6}{11}=\dfrac{33}{44}-\dfrac{24}{44}=\dfrac{9}{44}\)
\(a, 48.19+81.48\)
\(=49.(19+81)\)
\(=49.100=4900\)
\(b,23.134-34.23\)
\(=23.(134-34)\)
\(=23.100=2300\)
Bài3: a)(1989x1990+3978):(1992x1991-3984)=. (1989x1990+1989x2):(1992x1991-1992x2)=[1989x(1990+2)]:[1992x(1991-2)]=[1989x1992]:[1992x1989]=38620888:38620888=1