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a) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
= \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+\frac{1}{2}-\frac{36}{41}\)
= \(\frac{1}{2}-\left\{\frac{11}{24}+\frac{13}{24}\right\}-\left\{\frac{5}{41}+\frac{36}{41}\right\}\)
=\(\frac{1}{2}-\frac{24}{24}-\frac{41}{41}\)
=\(\frac{1}{2}-1-1\)
=\(\frac{-3}{2}\)
b) \(-12:\left\{\frac{3}{4}-\frac{5}{6}\right\}^2\)
= \(-12:\left\{\frac{9}{12}-\frac{10}{12}\right\}^2\)
= \(-12:\left\{\frac{-1}{12}\right\}^2\)
= \(-12:\frac{1}{144}\)
= \(-12.144\)
= -1728
c) \(\frac{7}{23}.\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left[\left(\frac{-24}{18}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-23}{6}\right)\)
= \(\frac{-7}{6}\)
d) \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)
= \(23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)
= \(\left\{23\frac{1}{4}-13\frac{1}{4}\right\}.\frac{7}{5}\)
= \(10.\frac{7}{5}\)
= 14
e) (1+23−14).(0,8−34)2
= (1+23−14).(\(\frac{4}{5}\)−34)2
= \(\left(\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right).\left(\frac{16}{20}-\frac{15}{20}\right)^2\)
= \(\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
= \(\frac{17}{20}.\frac{1}{400}\)
= \(\frac{17}{8000}\)
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