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A.-9/10*5/14*1/10*(-9/2)+1/7*(-9/10)
=-9/10*(5/14+1/7)*1/10*(-9/2)
=-9/10*1/2*1/10*(-9/2)
=-9/20*1/10*(-9/10)
=-9/200*-9/10
=81/2000
B. (1999/2011-2011/1999)-(-12/1999-12/2011)
=1999/2011-2011/1999+12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
= 1-1
=0
\(\frac{1999}{2011}-\frac{2011}{1999}-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\left(\frac{1999}{2011}+\frac{12}{2011}\right)-\frac{2011}{1999}+\frac{12}{1999}\)
\(=1-\left(\frac{2011}{1999}-\frac{12}{1999}\right)\)
\(=1-1\)
\(=0\)
NHÉ !!!!!!!
Ta có : \(\frac{1999}{2011}-\frac{2011}{1999}-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
=\(\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
=\(\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{12}{1999}-\frac{2011}{1999}\right)\)
=1-1
=0
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(-\frac{2011}{1999}+\frac{12}{1999}\right)\)
\(=\frac{2011}{2011}+\frac{-1999}{1999}=1+\left(-1\right)=0\)
(1999/2011-2011/1999) - (-12/1999 - 12/2011)
= 1999/2011 - 2011/1999 + 12/1999 + 12/2011
= (1999/2011 + 12/2011) - 1999/1999 - 12/1999 + 12/1999
= ( 1-1 ) + ( -12/1999 + 12/1999)
= 0 + 0
= 0
(1999/2011-2011/1999)-(12/1999-12/2011)
=1999/2011-2011/1999-12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
=1-1
=0
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(\frac{-12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(\frac{-2011}{1999}+\frac{12}{1999}\right)=\frac{2011}{2011}-\frac{1999}{1999}=1-1=0\)
\(\left(\frac{1999}{2011}-\frac{2011}{1999}\right)-\left(-\frac{12}{1999}-\frac{12}{2011}\right)\)
\(=\frac{1999}{2011}-\frac{2011}{1999}+\frac{12}{1999}+\frac{12}{2011}\)
\(=\left(\frac{1999}{2011}+\frac{12}{2011}\right)+\left(-\frac{2011}{1999}+\frac{12}{1999}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
A=1999/2011-2011/1999-(-12/1999)-12/2011
A = ( 1999/2011 - 2011/1999 ) - ( -12/1999 - 12/2011 )
=1999/2011-2011/1999+12/1999+12/2011
=(1999/2011+12/2011)-(2011/1999-12/1999)
=1-1
=0
Vậy A=0
B = 2/5 + [ 3/11 + (-2)/5 ]
=[2/5+(-2/5)]+3/11
=0+3/11
=3/11