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31.(17-3)-31.(17+3)
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a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
\(a)\)
\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)
\(=\left(-31\right)+50-19-150+31\)
\(=\left(-150\right)-19\)
\(=-169\)
\(b)\)
\(25.\left(45-17\right)-45.\left(25-17\right)\)
\(=25.45-25.17-45.25+45.17\)
\(=0\)
\(c)\)
\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)
\(d)\)
\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)
\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)
\(=1-2\)
\(=-1\)
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
\(a.\)
\(\dfrac{27}{13}-\dfrac{106}{111}+-\dfrac{5}{111}=\dfrac{27}{13}-\dfrac{106}{111}-\dfrac{5}{111}=\dfrac{27}{13}-\left(\dfrac{106+6}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
\(b.\)
\(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
\(c.\)
\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+-\dfrac{6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{25}{31}+\dfrac{6}{31}\right)=1-1=0\)
a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}=\dfrac{27}{13}+\left(\dfrac{-106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}+-1=\dfrac{14}{13}\)
b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\left(\dfrac{7}{19}+\dfrac{12}{19}\right)=\dfrac{12}{11}+1=\dfrac{23}{11}\)
c)\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{-25}{31}+\dfrac{-6}{31}\right)=1+-1=0\)
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
\(\dfrac{5}{3}\times\dfrac{3}{17}+\dfrac{3}{17}\times\dfrac{1}{3}-\dfrac{3}{17}\times2\)
\(=\dfrac{3}{17}\times\left(\dfrac{5}{3}+\dfrac{1}{3}+2\right)=\dfrac{3}{17}\times4=\dfrac{12}{17}\)
\(a,18.17-3.6.7\)
\(=18.17-18.7\)
\(=18.\left(17-7\right)\)
\(=18.10\)
\(=180\)
\(b,54-6.\left(17+9\right)\)
\(=54-6.17+6.9\)
\(=54-102+54\)
\(=\left(54+54\right)-102\)
\(=108-102\)
\(=6\)
\(c,33.\left(17-5\right)-17.\left(33-5\right)\)
\(=33.17-33.5-17.33-17.5\)
\(=\left(33.17-17.33\right)-\left[5\left(33-17\right)\right]\)
\(=0-5.16\)
\(=-80\)
Lời giải:
$=3.17+(3.120-3.17)=3.17+3.120-3.17$
$=(3.17-3.17)+3.120=3.120=360$
-3.(-17)+3.(120-17)
= 3.17 + 3.120 - 3.17
= (3.17-3.17) + 3.120
= 0 + 360 = 360
Lời giải:
$-3(-17)+3(120-17)=3.17+3.120-3.17=3.120=360$
31.(17-3)-31.(17+3)
=\(-31[-\left(17-3\right)-\left(17+3\right)|\)
= -31(-17+3-17-3)
=-31.(-34)
=1054
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