\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)

\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)

\(=26-6\dfrac{7}{32}\)

\(=20-\dfrac{7}{32}\)

\(=\dfrac{633}{32}\)

\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)

\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)

\(=26-6\dfrac{7}{32}\)

\(=20-\dfrac{7}{32}\)

\(=\dfrac{633}{32}\)

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

\(=\dfrac{7}{15}x\dfrac{5}{29}+\dfrac{13}{15}x\dfrac{7}{29}+\dfrac{11}{29}x\dfrac{7}{15}\)

\(=\left(\dfrac{7}{15}+\dfrac{13}{15}+\dfrac{7}{15}\right)x\left(\dfrac{5}{29}+\dfrac{7}{29}+\dfrac{11}{29}\right)=\dfrac{27}{15}x\dfrac{23}{29}=\dfrac{207}{5}\)

\(=\dfrac{7}{15}\times\dfrac{5}{29}+\dfrac{13}{15}\times\dfrac{7}{29}+\dfrac{11}{29}\times\dfrac{7}{15}\)

\(=\left(\dfrac{7}{15}+\dfrac{11}{15}+\dfrac{7}{15}\right)\times\left(\dfrac{5}{29}+\dfrac{7}{29}+\dfrac{11}{29}\right)=\dfrac{27}{15}\times\dfrac{23}{29}=\dfrac{207}{5}\)

`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

\(=200-100:5+1\)

\(=200-20+1\)

=\(180+1\)

\(=181\)

181

a)2486+(-486)-(-697)-(+97)+58

=2486-486+697-97+58

=(2486-486)+(697-97)+58

=2000+600+58

=2658

b)153.177+(-153).77

=153.177+153.(-77)

=153.(177-77)

=153.100=15300

c)-39.(5-99)+99.(10-39)

=-39.5-99+99.10-39

=-39.(5-10)+(-99+99)

=-39.(-5)+0

=195