\(a, \)\(\left(3^2\right)^2-\left(2^3\right)^2-\left(-5^2\right)^2=9^2-8^2-10^2\)

= \(81-64-100\)

\(=-83\)

\(b,\)\(2^3+3.\left(-\dfrac{1}{2}\right)^0-\left(\dfrac{1}{2}\right)^2.4+\left(\left(-2\right)^2:\dfrac{1}{2}\right):8=8+3.1-\dfrac{1}{4}.4+\left(4:\dfrac{1}{2}\right):8\) \(=8+3-1+8:8\)

\(=8+3-1+1\)

\(=11\)

a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)

\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)

b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)

\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)

c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

=1/4+3/4

=1

Quy luật có đúng ko vậy bạn

\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

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

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

a) S = 1.2 + 2.3 + 3.4 + ... + 99.100

S có thể được viết lại thành:

S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)

= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98

= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)

Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:

S = n(n+1)(2n+1)/6

Với n = 99, ta có:

S = 99.100.199/6 = 331650

Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:

S = n(n+1)/2

Với n = 98, ta có:

S = 98.99/2 = 4851

Do đó, S = 331650 - 4851 = 326799

b) B = 4924.12517.28−530.749.45529.162.748

B có thể được viết lại thành:

B = (4924.12517.28) / (530.749.45529.162.748)

B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)

B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529

B = 108 / 45529

c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101

C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101

C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)

C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)

C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)

C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)

d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018

D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^

A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))³

A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)

A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)

A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)

A = \(\dfrac{-73}{50}\)

B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))

B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)

B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)

B = \(\dfrac{13}{12}\)

\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(=1-\dfrac{1}{n^2+2n+1}\)

\(=\dfrac{n^2+2n}{n^2+2n+1}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Xét thừa số tổng quát:

\(\dfrac{k}{\left(\dfrac{k-1}{2}.\dfrac{k+1}{2}\right)^2}\)\(=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}\)

\(=\dfrac{k}{\dfrac{\left[\left(k-1\right)\left(k+1\right)\right]^2}{16}}=\dfrac{k}{\dfrac{\left(k^2-1\right)^2}{16}}=\dfrac{16k}{\left(k^2-1\right)^2}\)

Thay \(k=3;5;....2n+1\) ta được:

\(\dfrac{16.3}{\left(3^2-1\right)^2}+\dfrac{16.5}{\left(5^2-1\right)^2}+....+\dfrac{16.n}{\left(n^2-1\right)^2}\)

\(=16.\left(\dfrac{3}{\left(3^2-1\right)^2}+\dfrac{5}{\left(5^2-1\right)^2}+...+\dfrac{n}{\left(n^2-1\right)^2}\right)\)

\(=16.\left(\dfrac{3}{\left[\left(3-1\right)\left(3+1\right)\right]^2}+\dfrac{5}{\left[\left(5-1\right)\left(5+1\right)\right]^2}+...+\dfrac{n}{\left[\left(n-1\right)\left(n+1\right)\right]^2}\right)\)

\(=16.\left(\dfrac{3}{4.16}+\dfrac{5}{16.36}+...+\dfrac{n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{12}{4.16}+\dfrac{20}{16.36}+...+\dfrac{4n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{36}+...+\dfrac{1}{\left(n-1\right)^2}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2}{4\left(n+1\right)^2}-\dfrac{4}{4\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{\left(n+1\right)^2-4}{4\left(n+1\right)^2}\right)=\dfrac{4\left(n+1\right)^2-16}{4\left(n+1\right)^2}\)

\(=\dfrac{4\left[\left(n+1\right)^2-4\right]}{4\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-4}{\left(n+1\right)^2}\)

Chúc bạn học tốt!!!

\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(A=1-\dfrac{1}{n^2+2n+1}\)

\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)