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a, A = 3 2 . 5 2 . 4 3 : 2 3 . 3 2 . 2005 0
= 3 2 . 5 2 . 2 6 : 2 3 . 3 2 . 1
= 3 . 5 2 . 2 3 = 3.25.8 = 600
b, B = 194.12+6.437.2+3.369.4
= 194.12+437.12+369.12
= 12.(194+437+369)
= 12.1000 = 12000
c, C = 5 16 + 16 5 3 17 - 3 10 2 4 - 4 2
= 5 16 + 16 5 3 17 - 3 10 4 2 - 4 2
= 5 16 + 16 5 3 17 - 3 10 . 0 = 0
d, D = 5 2007 - 5 2006 : 5 2005 . 5
= 5 2007 - 5 2006 : 5 2006
= 5 2006 . 5 - 1 : 5 2006 = 4
\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)
\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)
\(=26-6\dfrac{7}{32}\)
\(=20-\dfrac{7}{32}\)
\(=\dfrac{633}{32}\)
\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)
\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)
\(=26-6\dfrac{7}{32}\)
\(=20-\dfrac{7}{32}\)
\(=\dfrac{633}{32}\)
Lời giải:
Gọi tổng trên là $A$
$A=2(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.11}+...+\frac{1}{100.103})$
$A=\frac{2}{3}(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{100.103})$
$=\frac{2}{3}(\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{103-100}{100.103})$
$=\frac{2}{3}(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103})$
$=\frac{2}{3}(1-\frac{1}{103})$
$=\frac{2}{3}.\frac{102}{103}=\frac{68}{103}$
Bạn Akai Haruma đáp án của bạn đúng khi phân số 1/7*11 là 1/7*10
\(=\dfrac{7}{15}x\dfrac{5}{29}+\dfrac{13}{15}x\dfrac{7}{29}+\dfrac{11}{29}x\dfrac{7}{15}\)
\(=\left(\dfrac{7}{15}+\dfrac{13}{15}+\dfrac{7}{15}\right)x\left(\dfrac{5}{29}+\dfrac{7}{29}+\dfrac{11}{29}\right)=\dfrac{27}{15}x\dfrac{23}{29}=\dfrac{207}{5}\)
`c) 5,6 . ( -45,39 ) - (-25,59) . 5,6 - 0,02 . 56`
`= 5,6 . (-45,39) + 25,59 . 5,6 - 0,2 . 5,6`
`= 5,6 . ( -45,39 + 25,59 - 0,2 )`
`= 5,6 . (-20)`
`= -112`
__________________________________________
`d) 5 / 17 . 9 / 23 + 9 / 23 . [-22] / 17`
`= 9 / 23 . ( 5 / 17 + [-22] / 17 )`
`= 9 / 23 . [-17] / 17`
`= [-9] / 23`
\(a,=28+19-28-32+57\\ =\left(28-28\right)+\left(19-32+57\right)\\ =0+44\\ =44\\ b,=39+13-26-62-39\\ =\left(39-39\right)+\left(13-26-62\right)\\ =0-75\\ =-75\\ c,=29+37+13+10-37-13\\ =\left(13-13\right)+\left(37-37\right)+\left(29+10\right)\\ =39\\ d,=-21-43-7-11+53+17\\ =\left(-21-11\right)+\left(-43+53\right)+\left(-7+17\right)\\ =-32+10+10\\ =-12\)
a,=28+19−28−32+57=(28−28)+(19−32+57)=0+44=44b,=39+13−26−62−39=(39−39)+(13−26−62)=0−75=−75c,=29+37+13+10−37−13=(13−13)+(37−37)+(29+10)=39d,=−21−43−7−11+53+17=(−21
g: \(\left(-196\right)-235+796-\left(-465\right)\)
\(=-196+796-235+465\)
\(=600+230=830\)
h: \(\left(-478\right)-\left(-192\right)+1478-192+465\)
\(=\left(-478+1478\right)+192-192+465\)
=1000+465
=1465
l: \(\left(-245\right)+4789-\left(-7545\right)-64789\)
\(=-245+4789+7545-64789\)
\(=\left(-245+7545\right)+\left(4789-64789\right)\)
\(=7300-60000=-52700\)
m: \(296-1985-2396+1985-744\)
\(=\left(296-2396\right)+\left(1985-1985\right)-744\)
=-2100-744
=-2844
n: \(1952+\left(-376\right)-7952-\left(-476\right)\)
\(=1952-7952-376+476\)
=100-6000
=-5900
o: \(\left(-475\right)-237-\left(-475\right)+9237+14\)
\(=\left(-475+475\right)+\left(-237+9237\right)+14\)
=9000+14
=9014