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\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\\ 25S=25\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 25S=1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\\ 25S-S=\left(1+\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2016}}\right)-\left(\dfrac{1}{5^2}+\dfrac{1}{5^4}+\dfrac{1}{5^6}+...+\dfrac{1}{5^{2018}}\right)\\ 24S=1-\dfrac{1}{5^{2018}}< 1\\ \Rightarrow S< \dfrac{1}{24}\)
2.
\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)
Vậy \(A< 3\)
b,
\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)
Vậy \(B< 2\)
\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)
\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)
\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)
\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)
Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)
Vậy P < Q
a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< x< \dfrac{-13}{5}:\dfrac{21}{15}=\dfrac{-13}{5}\cdot\dfrac{5}{7}=\dfrac{-13}{7}\)
=>-10<x<-13/7
hay \(x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< x< \dfrac{-2}{3}\cdot\dfrac{4-3-9}{12}\)
\(\Leftrightarrow-\dfrac{13}{9}< x< \dfrac{4}{9}\)
mà x là số nguyên
nên \(x\in\left\{-1;0\right\}\)
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. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
\(S=\dfrac{1}{5^2}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{2022}}\)
=>\(25\cdot S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}\)
=>\(25S-S=1+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2020}}-\dfrac{1}{5^2}-\dfrac{1}{5^4}-...-\dfrac{1}{5^{2022}}\)
=>\(24S=1-\dfrac{1}{5^{2022}}\)
=>\(S=\dfrac{1}{24}-\dfrac{1}{24\cdot5^{2022}}< \dfrac{1}{24}\)