a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

Vậy \(A=\dfrac{3}{11}\)

b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)

Vậy \(B=6\)

a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)

= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)

= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)

= \(\dfrac{3}{11}\)

b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)

= \(\dfrac{3}{5}\)

c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)

= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)

= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)

= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)

a)\(\frac{7}{12}.\frac{6}{11}+\frac{7}{12}.\frac{5}{11}-2\frac{7}{12}\)

\(=\frac{7}{12}.\left(\frac{6}{11}+\frac{5}{11}\right)-\frac{31}{12}\)

\(=\frac{7}{12}-\frac{31}{12}\)

\(=-2\)

b)\(\frac{-5}{9}.\frac{-6}{13}+\frac{5}{-9}.\frac{-5}{13}-\frac{5}{9}\)

\(=\frac{5}{9}.\left(\frac{6}{13}+\frac{5}{13}-1\right)\)

\(=\frac{5}{9}.\left(\frac{11}{13}-\frac{13}{13}\right)\)

\(=\frac{5}{9}.\frac{-2}{13}\)

\(=-\frac{10}{117}\)

c)\(0,8.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-1\frac{2}{5}\)

\(=\frac{4}{5}.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-\frac{15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)

\(=\frac{4}{5}.\left(-2\right)-\frac{7}{5}\)

\(=\frac{-8}{5}-\frac{7}{5}\)

\(=-3\)

d)\(-75\%.\frac{6}{7}+5\%.\frac{6}{7}+\frac{7}{10}.1\frac{1}{7}\)

\(=\frac{-15}{20}.\frac{6}{7}+\frac{1}{20}.\frac{6}{7}+\frac{7}{10}.\frac{8}{7}\)

\(=\frac{6}{7}.\left(\frac{-15}{20}+\frac{1}{20}\right)+\frac{4}{5}\)

\(=\frac{6}{7}.\frac{-7}{10}+\frac{4}{5}\)

\(=-\frac{3}{5}+\frac{4}{5}\)

\(=\frac{1}{5}\)

Linz

a,\(\dfrac{5}{6}+\left(-\dfrac{1}{2}\right)+\dfrac{3}{4}\)

\(=\dfrac{10}{12}+\left(-\dfrac{6}{12}\right)+\dfrac{9}{12}\)

\(=\dfrac{10-6+9}{12}=\dfrac{13}{12}\)

b,\(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)

\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right):\dfrac{7}{15}\)

\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right):\dfrac{7}{15}\)

\(=\dfrac{5}{12}:\dfrac{7}{15}\)

\(=\dfrac{25}{28}\)

c,\(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)

\(=\dfrac{7}{12}-\dfrac{5}{8}\)

\(=\dfrac{14}{24}-\dfrac{15}{24}\)

\(=-\dfrac{1}{24}\)

d,\(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)

\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)

\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)

\(=\dfrac{49}{12}.\dfrac{12}{13}\)

\(=\dfrac{49}{13}\)

a) \(\dfrac{5}{6}+\left(\dfrac{-1}{2}\right)+\dfrac{3}{4}\)

\(=\dfrac{10}{12}-\dfrac{6}{12}+\dfrac{9}{12}\)

\(=\dfrac{13}{12}\)

b) \(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)

\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right).\dfrac{15}{7}\)

\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right).\dfrac{15}{7}\)

\(=\dfrac{5}{12}.\dfrac{15}{7}\)

\(=\dfrac{25}{28}\)

c) \(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)

\(=\dfrac{7}{12}-\dfrac{5}{8}\)

\(=\dfrac{14}{24}-\dfrac{15}{24}\)

\(=\dfrac{-1}{24}\)

d) \(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)

\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)

\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)

\(=\dfrac{49}{12}.\dfrac{12}{13}\)

\(=\dfrac{49}{13}\)

ngu như con bò tót, ko biết 1+1=2.

a, \(\frac{1}{4}+\frac{5}{12}-\frac{1}{13}-\frac{7}{8}\)

\(=\left(\frac{1}{4}+\frac{5}{12}\right)-\left(\frac{1}{13}+\frac{7}{8}\right)\)

\(=\frac{2}{3}-\frac{99}{104}\)

\(=-\frac{89}{312}\)

b, \(11\frac{3}{13}-2\frac{4}{7}+5\frac{3}{13}\)

\(=\left(11\frac{3}{13}+5\frac{3}{13}\right)-2\frac{4}{7}\)

\(=\frac{214}{13}-\frac{18}{7}\)

\(=\frac{1264}{91}\)

c, \(\left(6\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)

\(=6\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)

\(=\left(6\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)

\(=2+3\frac{7}{11}\)

\(=5\frac{7}{11}\)

\(=\frac{62}{11}\)

d, \(\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-0,25-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

\(=\left(6,17+3\frac{5}{9}-2\frac{36}{97}\right)\cdot0\)

\(=0\)

e, \(-1,5\cdot\left(1+\frac{2}{3}\right)\)

\(=-\frac{3}{2}\cdot\frac{5}{3}\)

\(=-\frac{5}{2}\)

f, Đặt \(A=1^2+2^2+3^2+...+100^2\)

\(=1+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)

\(=1+2\cdot3-2+3\cdot4-3+...+100\cdot101-100\)

\(=\left(2\cdot3+3\cdot4+...+100\cdot101\right)-\left(1+2+3+...+100\right)\)

Đặt B = 2 . 3 + 3 . 4 + ... + 100 . 101 

3B = 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 100 . 101 . ( 102 - 99 )

3B = 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... + 100 . 101 . 102 - 99 . 100 . 101 

3B = 100 . 101 . 102

B = \(\frac{100\cdot101\cdot102}{3}\)

B = 343400

Thay B vào A. Ta được :

\(A=343400-\left(1+2+3+...+100\right)\)

Thay C = 1 + 2 + 3 + ... + 100

Dãy số 1; 2; 3; ...; 100 có số số hạng là:

( 100 - 1 ) : 1 + 1 = 100 ( số hạng )

Tổng của dãy số đó là :

( 100 + 1 ) . 100 : 2 = 5050

=> C = 5050

Thay C vào A. Ta được :

\(A=343400-5050\)

\(A=338350\)

Vậy A = 338350