\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{1990^2}\)
C/M : \(M< \dfrac{3}{4}\)
Những câu hỏi liên quan
Chứng tỏ:
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1995}-\dfrac{1}{1996}=\dfrac{1}{996}+\dfrac{1}{997}+...+\dfrac{1}{1990}\)
Giúp Mình vs
\(2+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x.\left(x+1\right)}=1\dfrac{1989}{1990}\)
\(2+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{2.\left(x+1\right)}=1\dfrac{1989}{1990}\)
làm nhanh nhé
1. Cho Ndfrac{1}{31}+dfrac{1}{32}+...+dfrac{1}{60}CMR dfrac{3}{5} N dfrac{4}{5}2. Cho Mdfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+...+dfrac{29}{3^{29}}-dfrac{30}{3^{30}}CMR M dfrac{3}{16}3. Cho Qdfrac{2}{3}+dfrac{8}{9}+dfrac{26}{27}+...+dfrac{3^{2021}-1}{3^{2021}}CMR Qdfrac{4041}{2}
Đọc tiếp
1. Cho N=\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\)
CMR \(\dfrac{3}{5}< N< \dfrac{4}{5}\)
2. Cho M=\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{29}{3^{29}}-\dfrac{30}{3^{30}}\)
CMR \(M< \dfrac{3}{16}\)
3. Cho Q=\(\dfrac{2}{3}+\dfrac{8}{9}+\dfrac{26}{27}+...+\dfrac{3^{2021}-1}{3^{2021}}\)
CMR \(Q>\dfrac{4041}{2}\)
M= \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+\(\dfrac{1}{1+2+3+...+59}\) CMR M<\(\dfrac{2}{3}\)
Ta có \(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\)
= \(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\)
= \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\dfrac{1}{3}-\dfrac{1}{60}=\dfrac{19}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)
Vậy M < \(\dfrac{2}{3}\)
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16 tháng 10 2021 lúc 18:40
Câu 1 : Rút gọn
\(G=\dfrac{6!}{\left(m-2\right)\left(m-3\right)}.\left[\dfrac{\left(m+1\right)!}{5!.\left(m-4\right)!.\left(m+1\right)}-\dfrac{m!}{12.3!.\left(m-4\right)!}\right]\)
Câu 2 : CMR
\(1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{n!}< 3\forall n\in N\)
2 tháng 7 2021 lúc 21:10
Tính giá trị biểu thức A , biết rằng A M : N Mà M dfrac{dfrac{1}{99}+dfrac{2}{98}+dfrac{3}{97}+dfrac{4}{96}+...+dfrac{97}{3}+dfrac{98}{2}+dfrac{99}{1}}{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+dfrac{1}{5}+dfrac{1}{6}+...+dfrac{1}{100}} N dfrac{92-dfrac{1}{9}-dfrac{2}{10}-dfrac{3}{11}-...-dfrac{90}{98}-dfrac{91}{99}-dfrac{92}{100}}{dfrac{1}{45}+dfrac{1}{50}+dfrac{1}{55}+...+dfrac{1}{495}+dfrac{1}{500}}
Đọc tiếp
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)
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tìm x,y viết dưới dạng phân số
a. 5+dfrac{x}{5+dfrac{2}{5+dfrac{3}{5+dfrac{4}{5}}}}dfrac{x}{1+dfrac{5}{2+dfrac{4}{3+dfrac{3}{5+dfrac{1}{6}}}}}
b.
dfrac{y}{3+dfrac{5}{2+dfrac{4}{2+dfrac{5}{2+dfrac{4}{2+dfrac{5}{3}}}}}}+dfrac{y}{7+dfrac{1}{3+dfrac{1}{3+dfrac{1}{4}}}}
2
c,
x.left(dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+dfrac{1}{1+1}}}}}}}}right)2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2+dfrac{1}{2}}}}}}}+x.left(3+dfrac{1}{3-df...
Đọc tiếp
tìm x,y viết dưới dạng phân số
a. \(5+\dfrac{x}{5+\dfrac{2}{5+\dfrac{3}{5+\dfrac{4}{5}}}}=\dfrac{x}{1+\dfrac{5}{2+\dfrac{4}{3+\dfrac{3}{5+\dfrac{1}{6}}}}}\)
b.
\(\dfrac{y}{3+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{2+\dfrac{4}{2+\dfrac{5}{3}}}}}}+\dfrac{y}{7+\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{4}}}}\)
= 2
c,
\(x.\left(\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+1}}}}}}}}\right)=\)\(2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2}}}}}}}\)+\(x.\left(3+\dfrac{1}{3-\dfrac{1}{3+\dfrac{1}{3+\dfrac{1}{3-\dfrac{1}{3}}}}}\right)\)
Giair bằng máy tính casio
bài này đúng là thị của phi...vô của lí ... :))
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1,CMR:\(1-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{1990}=\dfrac{1}{996}+\dfrac{1}{997}+\dfrac{1}{1990}\)
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+...+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\)
c/m A<\(\dfrac{3}{16}\)
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(4A=A+3A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(12A=3-1+\frac{1}{3}-\frac{1}{3^2}+....-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow 16A=12A+4A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}<3\)
\(\Rightarrow A< \frac{3}{16}\)
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