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Lời giải:
$167(222-345)+345(167+1)-167.345$
$=167.222-167.345+345.167+345-167.345$
$=167.222+(-167.345+167.345)+345-167.345$
$=167.222+345-167.345$
$=167(222-345)+345=167.(-123)+345=-20196$
\(\left(\frac{3}{17}-\frac{2}{345}+\frac{5}{12}\right)-\left(-\frac{2}{345}+\frac{3}{17}-\frac{1}{12}\right)\)
\(=\frac{3}{17}-\frac{2}{345}+\frac{5}{12}+\frac{2}{345}-\frac{3}{17}+\frac{1}{12}\)
\(=\left(\frac{3}{17}-\frac{3}{17}\right)+\left(\frac{2}{345}-\frac{2}{345}\right)+\frac{5}{12}+\frac{1}{12}\)
\(=\frac{6}{12}=\frac{1}{2}\)
( 3/17 - 2/345 + 5/12 ) - ( - 2/345 + 3/17 - 1/12 )
= 3/17 - 2/345 + 5/12 + 2/345 - 3/17 + 1/12
= ( 3/17 - 3/17 ) + ( 2/345 - 2/345 ) + ( 5/12 + 1/12 )
= 0 + 0 + 1/2
=1/2
a/ -137.45+ (-45).(-37)
= -6165 + 1665
= - 4500
b/(-18+98):5-(-4)^2
=80 : 5 - (-16)
= 16 - (-16)
= 16 + 16
= 32
c/ (578-950)-(-950+578-23)
= 578 - 950 - 950 - 578 + 23
= { 578 - 578 } - {950 - 950} + 23
= 0 - 0 + 23
= 0 + 23
= 23
d/9-x : (-2) = -15
9 - x = -15 . (-2)
9 - x = 30
x = 9 - 30
x = -21
\(1-\frac{385}{386}=\frac{1}{386}\)
\(1-\frac{578}{579}=\frac{1}{579}\)
Ta có: \(\frac{1}{386}>\frac{1}{579}\)
\(\Rightarrow1-\frac{578}{579}< 1-\frac{385}{386}\)
\(\Rightarrow-\frac{578}{579}< -\frac{385}{386}\)
Vậy \(-\frac{578}{579}< -\frac{385}{386}\)
(123+345) + ( 456- 123)- [2017 -(-345)]
=123+345+456-123-2017-345
=(123-123)+(345-345)+(456-2017)
=0+0-1561
=-1561
345-578+[-1]=345+[-578]+[-1]
=-233+[-1]
=-234