Tính bằng cách hợp lí
a) 1 3 . 2 5 + 1 3 . 4 5 − 1 5 . 1 3
b) − 5 9 . 5 8 + 3 9 . − 5 8 + 5 9
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2:
a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)
1:
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)
Bài 1:
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)
\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)
\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)
\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)
\(A=\dfrac{-16-1}{4}\)
\(A=-\dfrac{17}{4}\)
Bài 2:
\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)
\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)
\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)
\(=\dfrac{1}{3}\cdot-2\)
\(=-\dfrac{2}{3}\)
a: =382-282+531-331
=100+200=300
b: =(7-8)+(9-10)+...+(2009-2010)
=(-1)+(-1)+....+(-1)
=-1002
c: =-(1+2+3+...+2009+2010)
=-2010*2011/2=-2021055
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
`a, 4/9 + 2/5 + 5/9= (4/9 +5/9) + 2/5= 1 + 2/5= 5/5 + 2/5= 7/5`
`b, 6/13 + 5/8 + 7/13=(6/13 + 7/13) + 5/8=1+5/8= 8/8 + 5/8= 13/8`
`c, 5/11 + 6/25 + 66/121 = ( 5/11 + 66/121 ) + 6/25=1 + 6/25 = 25/25 + 6/25 = 31/25`
`d, 1/2 + 1/3 + 1/4 + 1/6 + 1/12 + 1/24= (1/2 + 1/4) +(1/3 + 1/6) + (1/12 + 1/24)= 3/4 + 1/2 + 1/8= 6/8 + 4/8 + 1/8= 11/8`
`@ yl`
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
`1 - 2 + 3 - 4 + 5 - 6 +...+ 2021 - 2022`
`= (1 - 2) + (3 - 4) + (5 - 6) +...+ (2021 - 2022)`
`= (-1) + (-1) + (-1) + ... + (-1) ` [có `2022 : 2 = 1011` nhóm]
`= (-1) xx 1011 = -1011`
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
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) 1 3 . 2 5 + 1 3 . 4 5 − 1 5 . 1 3 = 1 3 . 2 5 + 4 5 − 1 5 = 1 3 . 5 5 = 1 3 .
b) − 5 9 . 5 8 + 3 9 . − 5 8 + 5 9 = − 5 9 . 5 8 + − 5 9 . 3 8 + -5 9 . − 1 = − 5 9 . 5 8 + 3 8 − 1 = − 5 9 . 8 8 − 1 = − 5 9 .0 = 0.