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a) (-2).3 + (-4) - 7.0 + 1
=-6+(-4)-0+1
=-10-0+1
=-10+1
=-9
b) (-1).(-2) + (-3).(-4) - (-2).(-3)
=2+12-6
=14-6
=8
c)(-1).(-2).(-3).(-4).(-5) : [(-3) – (-5)]
=2.(-3).(-4).(-5):[(-3)+5]
=-6.(-4).(-5):2
=24.(-5):2
=-120:2
=-60
nhớ k giùm mình nhaa~
4:
a: =4/15-2,9+11/15=1-2,9=-1,9
b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)
c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)
d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)
e: =-5/12-7/12-3,7-6,3=-1-10=-11
f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10
\(a,\dfrac{1}{3}-\left(-1\dfrac{2}{5}\right)+\left(-3\dfrac{1}{4}\right)\\ =\dfrac{1}{3}-\left(-\dfrac{7}{5}\right)-\dfrac{13}{4}\\ =\dfrac{1}{3}+\dfrac{7}{5}-\dfrac{13}{4}\\ =\dfrac{20}{3\times20}+\dfrac{7\times12}{5\times12}-\dfrac{13\times15}{4\times15}\\ =\dfrac{20+84-195}{60}\\ =\dfrac{-91}{60}\)
\(b,\dfrac{5}{4}-\left(-3\dfrac{1}{2}\right)-\dfrac{7}{10}\\ =\dfrac{5}{4}+\dfrac{7}{2}-\dfrac{7}{10}\\ =\dfrac{5\times5}{4\times5}+\dfrac{7\times10}{2\times10}-\dfrac{7\times2}{10\times2}\\ =\dfrac{25+70-14}{20}\\ =\dfrac{81}{20}\)
\(c,\dfrac{3}{2}-\left[\left(-\dfrac{4}{7}-\left(\dfrac{1}{2}+\dfrac{5}{8}\right)\right)\right]\\ =\dfrac{3}{2}-\left[-\dfrac{4}{7}-\left(\dfrac{4}{8}+\dfrac{5}{8}\right)\right]\\ =\dfrac{3}{2}-\left(-\dfrac{4}{7}-\dfrac{9}{8}\right)\\ =\dfrac{3}{2}+\dfrac{4}{7}+\dfrac{9}{8}\\ =\dfrac{3\times28}{2\times28}+\dfrac{4\times8}{8\times7}+\dfrac{9\times7}{8\times7}\\ =\dfrac{84+32+63}{56}\\ =\dfrac{179}{56}\)
`@` `\text {Ans}`
`\downarrow`
`a,`
\(\dfrac{1}{3}-\left(-1\dfrac{2}{5}\right)+\left(-3\dfrac{1}{4}\right)\)
`= 1/3+7/5 - 13/4`
`= 26/15 - 13/4`
`= -91/60`
`b,`
\(\dfrac{5}{4}-\left(-3\dfrac{1}{2}\right)-\dfrac{7}{10}\)
`= 5/4+7/2 - 7/10`
`= 1,25 + 3,5 - 0,7`
`= 4,75 - 0,7`
`= 4,05`
`c,`
\(\dfrac{3}{2}-\left[\left(-\dfrac{4}{7}\right)-\left(\dfrac{1}{2}+\dfrac{5}{8}\right)\right]\)
`= 3/2 - [(-4/7) - 9/8]`
`= 3/2 - (-95/56)`
`= 179/56`
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a) \(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{6-3+2}{6}=\dfrac{1}{6}\)
\(b.\) \(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{6+10}{15}=\dfrac{16}{15}\)
\(c.\) \(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{4}{11}=\dfrac{21}{44}+\dfrac{7}{44}+\dfrac{16}{44}=\dfrac{21+7+16}{44}=\dfrac{44}{44}=1\)
a/\(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{6}{6}-\dfrac{3}{6}+\dfrac{2}{6}=\dfrac{5}{6}\)
b/\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}=\dfrac{2}{5}+\dfrac{3}{5}.\dfrac{10}{9}=\dfrac{2}{5}+\dfrac{2}{3}=\dfrac{6}{15}+\dfrac{10}{15}=\dfrac{16}{15}\)
a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
Giải:
a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\)
=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\)
=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\)
=\(\dfrac{7}{5}+\dfrac{-1}{4}\)
=\(\dfrac{23}{20}\)
b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\)
=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\)
=\(\left[0+0\right]:\dfrac{2017}{2018}\)
=0\(:\dfrac{2017}{2018}\)
=0
c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)
=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10
Bài 2:
a: A=(12,87+14,13)+(-14,7-37,3)
=27-52=-25
b: B=-1/3+2/5-2/3-3/5+1/5
=-1
a, [ (- 1) . 3 ] + (- 4) - (7.0) + 1
= [- 3 + ( - 4 )] - 0 + 1
= - 7 - 0 + 1
= - 6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)
= [(-1).(-2)] + [(-3).(-4)] - [(-2).(-3)]
= 2 + 12 - 6
= 8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]
= (-1).(-2).(-3).(-4).(-5) : 2
= - 120 : 2
= - 60
Chắc chắn đấy!
a, (-1).3+(-4)-7.0+1=-3-4-0+1=-6
b, (-1).(-2)+(-3).(-4)-(-2).(-3)=2+12-6=8
c, (-1).(-2).(-3).(-4).(-5):[(-3)-(-5)]=(-120):2=-60