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Lời giải:
$=3.17+(3.120-3.17)=3.17+3.120-3.17$
$=(3.17-3.17)+3.120=3.120=360$
a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
`#3107.101107`
\(-16 \div (-8)+5[3-15\div5+2(-3+4)]\)
`= 2 + 5(3 - 3 - 6 + 8)`
`= 2 + 5*(-6 + 8)`
`= 2 + 5*2`
`= 2(5 + 1)`
`= 2*6 = 12`
____
`15*(-236)+15*235`
`= 15 * (-236 + 235)`
`= 15 * (-1)`
`= -15`
____
`237*(-28)+28*137`
`= 28 * (-237 + 137)`
`= 28 * (-100)`
`= -2800`
____
`(-3)*(-17)+3*(120-17)`
`= 3 * 17 + 3 * (120 - 17)`
`= 3 * (17 + 120 - 17)`
`= 3 * 120`
`= 360`
____
`(-8)*72+8*(-19)-(-8)`
`= 8 * (-72) + 8 * (-19) + 8`
`= 8 * (-72 - 19 + 1)`
`= 8 * (-90)`
`= -720`
a)−16÷(−8)+5[3−15÷5+2(−3+4)]
=2+5(3−3−6+8)
=2+5⋅(−6+8)
=2+5⋅2
=2(5+1)
=2⋅6
=12
b)15⋅(−236)+15⋅235
=15⋅(−236+235)
=15⋅(−1)
=−15
c)237⋅(−28)+28⋅137
=28⋅(−237+137)
=28⋅(−100)
=−2800
d)(−3)⋅(−17)+3⋅(120−17)
=3⋅17+3⋅(120−17)
=3⋅(17+120−17)
=3⋅120
=360
e)(−8)⋅72+8⋅(−19)−(−8)
=8⋅(−72)+8⋅(−19)+8
=8⋅(−72−19+1)
=8⋅(−90)
=−720
\(a)\)
\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)
\(=\left(-31\right)+50-19-150+31\)
\(=\left(-150\right)-19\)
\(=-169\)
\(b)\)
\(25.\left(45-17\right)-45.\left(25-17\right)\)
\(=25.45-25.17-45.25+45.17\)
\(=0\)
\(c)\)
\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)
\(d)\)
\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)
\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)
\(=1-2\)
\(=-1\)
a) (-31)+(50-19)-(150-31)
= (-31)+50+(-19)-150+(-31)
= (-31)+50-150+(-19)-(-31)
= (-31)+(-100)+12
= -119
b) 25(45-17)-45(25-17)
= 25.45-25.17-45.25-45.17
= 25(45-45)-25(17-17)
= 0
c) -1/12 + 4/3
= -1/12 + 16/12
= 15/12
= 5/4
d) 3/5+(-5)/20+30/75+(-7)/4
= 45/75+30/75+(-5)/20+(-35)/20
= 1+(-2)
= -1
\(\frac{17}{33}.\frac{2}{5}+\frac{3}{5}.\frac{17}{33}-\frac{17}{33}\)
\(=\frac{17}{33}\times\frac{2}{5}+\frac{3}{5}\times\frac{17}{33}-\frac{17}{33}\times1\)
\(=\frac{17}{33}\times\left(\frac{2}{5}+\frac{3}{5}-1\right)\)
\(=\frac{17}{33}\times\left(1-1\right)\)
\(=\frac{17}{33}\times0=\frac{17}{33}\)
a) 4.(1 930 + 2 019) + 4.(-2 019)
= 4.(1 930 + 2 019 - 2 019)
= 4.1 930
= 7 720
b) (-3).(-17) + 3.(120 - 17)
= 3.17 + 3.(120 - 17)
= 3.(17 + 120 - 17)
= 3.120
= 360
-3.(-17)+3.(120-17)
= 3.17 + 3.120 - 3.17
= (3.17-3.17) + 3.120
= 0 + 360 = 360
Lời giải:
$-3(-17)+3(120-17)=3.17+3.120-3.17=3.120=360$