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Giải:
1) (-8/13:3/7+-5/13:3/7).(-4)3.|-3|/7
=[7/3.(-8/13+-5/13)].-192/7
=[7/3.(-1)].-192/7
=-7/3.-192/7
=64
2) 75%-(5/2+5/3)+(-1/2)2
=3/4-25/6+1/4
=(3/4+1/4)-25/6
=1-25/6
=-19/6
Chúc bạn học tốt!
1) \(\left(\dfrac{-8}{13}:\dfrac{3}{7}+\dfrac{-5}{13}:\dfrac{3}{7}\right).\dfrac{\left(-4\right).|-3|}{7}\)
= \(\left[\left(\dfrac{-8}{13}+\dfrac{-5}{13}\right):\dfrac{3}{7}\right].\dfrac{-64.3}{7}\)
= \(\left[-1:\dfrac{3}{7}\right].\dfrac{-192}{7}\)
= \(\dfrac{-7}{3}.\dfrac{-192}{7}\)
= \(64\)
2) \(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(-\dfrac{1}{2}\right)^2\)
= \(\dfrac{3}{4}-\dfrac{25}{6}+\dfrac{1}{4}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)-\dfrac{25}{6}\)
= \(1-\dfrac{25}{6}\)
= \(\dfrac{-19}{6}\)
Chúc bạn học tốt !
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
1, \(\dfrac{-3}{5}:\dfrac{7}{5}-\dfrac{3}{5}:\dfrac{7}{5}+2\dfrac{3}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{3}{5}\right):\dfrac{7}{5}+\dfrac{13}{5}\)
\(=\dfrac{-6}{5}:\dfrac{7}{5}+\dfrac{13}{5}\)
\(=\dfrac{-6}{7}+\dfrac{13}{5}\\ =\dfrac{61}{35}\)
2, \(0,75-\left(2\dfrac{1}{3}+0,75\right)+3^2\cdot\left(-\dfrac{1}{9}\right)\)
\(=0,75-\dfrac{37}{12}+9\cdot\left(-\dfrac{1}{9}\right)\)
\(=-\dfrac{7}{3}+\left(-1\right)\\ -\dfrac{10}{3}\)
a: =11+3/4-6-5/6+4+1/2+1+2/3
=10+9/12-10/12+6/12+8/12
=10+13/12=133/12
b: \(=2+\dfrac{17}{20}-1-\dfrac{11}{15}+2+\dfrac{3}{20}\)
=3-11/15
=34/15
c: \(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}=\dfrac{5}{7}\)
d: \(=\dfrac{29}{8}\cdot\dfrac{36}{29}\cdot\dfrac{15}{23}\cdot\dfrac{23}{5}=\dfrac{9}{2}\cdot3=\dfrac{27}{2}\)
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
1, \(\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}=\left(\dfrac{2}{3}-\dfrac{3}{2}\right)\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{6}\cdot\dfrac{3}{4}+\dfrac{1}{2}=-\dfrac{5}{8}+\dfrac{1}{2}=-\dfrac{1}{8}\)
2, \(\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=\left(-1\right)+1-\dfrac{3}{7}=0-\dfrac{3}{7}=-\dfrac{3}{7}\)
\(\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}=\dfrac{-5}{6}x\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{-5}{8}+\dfrac{1}{2}=\dfrac{-1}{8}\)
\(\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}=-1+1-\dfrac{3}{7}=0-\dfrac{3}{7}=\dfrac{-3}{7}\)
\(75\%-\left(\dfrac{5}{2}+\dfrac{5}{3}\right)+\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{5}{3}+\dfrac{1}{4}\)
\(=\dfrac{9}{12}-\dfrac{30}{12}-\dfrac{20}{12}+\dfrac{3}{12}\)
\(=\dfrac{-38}{12}=\dfrac{-19}{6}\)