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d. 12.37 + 12.78 + 115.29 + 115.59
= 12 . ( 37 + 78 ) + 115 . ( 29 + 59 )
= 12 . 115 + 115 . 88
= ( 12 + 88 ) . 115
= 100 . 115
= 11 500
a. 45.34 + 45.37 + 71.23 + 33.71
= 45 . ( 34 + 37 ) + 71 . ( 23 + 33 )
= 45 . 71 + 71 . 56
= ( 45 + 56 ) . 71
= 101 . 71
= ( 100 + 1 ) . 71
= 7100 + 71
= 7171
\(53,68-37.63-12.37\)
\(=53,68-\left(37.63+12.37\right)\)
\(=53,68-\left[37\left(63+12\right)\right]\)
\(=53,68-\left[37.75\right]\)
\(=53,68-2775\)
\(=\text{-2721.32}\)
53,68-37.63-12.37=53,68-[37.(63+12)]
=53,68-(37.75)
=53,68-2775
=-2721.32
( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 1 + 1/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x ( 4/3 - 4/3)
=( 2013 x 2014 + 2014 x 2015 + 2015 x 2016) x 0
=0
Ta có: \(\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(=\left(2013\cdot2014+2014\cdot2015+2015\cdot2016\right)\left(\dfrac{3}{3}+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
=0
\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)
\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)
a,50%+127−21=21+127−21=(21−21)+127=127b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12
a. 45.34 + 45.37 + 71.23 + 33.71
= 45.(34 + 37) + 71.(23 + 33)
= 45.71 + 71.56
= 71.(45 + 56)
= 71 . 101
= 7171
a) \(45.34+45.37+71.23+33.71=45\left(34+37\right)+71\left(23+33\right)=45.71+71.56=71\left(45+56\right)=71.101=7171\)
b) \(23.46+23.52+98.35+98.42=23\left(46+52\right)+98\left(35+42\right)=23.98+98.77=98\left(23+77\right)=98.100=9800\)
c) \(27.34+27.56+90.20+26.90=27\left(34+56\right)+90\left(20+26\right)=27.90+90.46=90\left(27+46\right)=90.73=6570\)
d) \(12.37+12.78+115.29+115.59=12\left(37+78\right)+115\left(29+59\right)=115.12+115.88=115\left(12+88\right)=115.100=11500\)