(-3,75).(-7,2)+2,8.3,75

<=>3,75.7,2+2,8.3,75   (đổi 2 dấu  âm thành dấu  dương)

= 3,75. (7,2+2,8)

=3,75 .10

=37,5

=> 3,75.7,2+2,8.3,75

=3,75(7,2+2,8)

=3,75.10

=37,5

=37,5

\(\left(-3,75\right).\left(-7,2\right)+2,8.3,75\)

\(=3,75.7,2+2,8.3,75\)

\(=3,75\left(7,2+2,8\right)\)

\(=3,75.10\)

\(=37,5\)

\(\left(-3,75\right)\left(-7,2\right)+2,8.3,75=3,75.7,2+2,8.3,75=\dfrac{15}{4}.\dfrac{36}{5}+\dfrac{14}{5}.\dfrac{15}{4}=\dfrac{15}{4}\left(\dfrac{36}{5}+\dfrac{14}{5}\right)=\dfrac{15}{4}.10=\dfrac{75}{2}=37.5\)

a) \(9\cdot3^2\cdot\dfrac{1}{81}=\dfrac{9\cdot9}{81}=\dfrac{81}{81}=1\)

b) \(2\dfrac{1}{2}+\dfrac{4}{7}:\left(-\dfrac{8}{9}\right)\)\(=\dfrac{5}{2}+\dfrac{4}{7}\cdot\left(-\dfrac{9}{8}\right)\)

\(=\dfrac{5}{2}-\dfrac{9}{14}\)\(=\dfrac{35}{14}-\dfrac{9}{14}=\dfrac{26}{14}=\dfrac{13}{7}\)

c) \(3,75\cdot7,2+2,8\cdot3,75=3,75\left(7,2+2,8\right)=3,75\cdot10=37,5\)

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

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

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

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

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

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

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

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

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

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

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