ta có :

99³+1+3(99²+99)

=99³+1+3.99²+3.99

=99³+3.99².1+3.99.1²+1³

= (99+1)³

\(47\left(23+50\right)-23\left(47+50\right)=47.23+47.50-23.47-23.50=50\left(47-23\right)=1200\)

\(85\left(35-47\right)-35\left(85-27\right)=85.35-85.47-85.35+35.27=35.27-85.47=-3050\)

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

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

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

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

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

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

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

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

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

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

b) = (-461) - 363

= - 824

d) = (-1010) + 1000

= -10

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