Cho \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
Chứng minh rằng: \(\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}=\dfrac{1}{\left(a+b+c\right)^{2n+1}}\)
Những câu hỏi liên quan
Áp dụng chứng minh rằng nếu : dfrac{1}{a}+dfrac{1}{b}+dfrac{1}{C}dfrac{1}{a+b+c}, thì:
dfrac{1}{a^{2n+1}}+dfrac{1}{b^{text{2n+1}}}+dfrac{1}{c^{text{2n+1}}}dfrac{1}{a^{text{2n+1}}+b^{text{2n+1}}+c^{text{2n+1}}}với nin N
giúp mk với...
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Áp dụng chứng minh rằng nếu : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{C}=\dfrac{1}{a+b+c}\), thì:
\(\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{\text{2n+1}}}+\dfrac{1}{c^{\text{2n+1}}}=\dfrac{1}{a^{\text{2n+1}}+b^{\text{2n+1}}+c^{\text{2n+1}}}\)với \(n\in N\)
giúp mk với...
20 tháng 12 2022 lúc 20:42
Cho:
\(A=\dfrac{1}{1.\left(2n-1\right)}+\dfrac{1}{3.\left(2n-3\right)}+...+\dfrac{1}{\left(2n-3\right).3}+\dfrac{1}{\left(2n-1\right).1}\) \(B=1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\) (với n ∈ N*).
Tính \(\dfrac{A}{B}\)
13 tháng 4 2023 lúc 18:48
Tính tổng: a) \(S=2C^2_{2n}+4C^4_{2n}+6C^6_{2n}+...+2nC^{2n}_{2n}\)
b) \(S=\dfrac{1}{2}C^0_{2n}+\dfrac{1}{4}C^2_{2n}+\dfrac{1}{6}C^4_{2n}+...+\dfrac{1}{2n+2}C^{2n}_{2n}\)
Cho A = \(\dfrac{1}{1.\left(2n-1\right)}+\dfrac{1}{3.\left(2n-3\right)}+...+\dfrac{1}{3.\left(2n-3\right)}+\dfrac{1}{1.\left(2n-1\right)}\); B = \(1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\). Tính \(\dfrac{A}{B}\)
cho
A=\(\dfrac{1}{1\left(2n-1\right)}+\dfrac{1}{3\left(2n-3\right)}+...+\dfrac{1}{\left(2n-3\right)3}+\dfrac{1}{\left(2n-1\right)1}\)
B=\(1+\dfrac{1}{3}+...+\dfrac{1}{2n-1}\)
tính \(\dfrac{A}{B}\)
16 tháng 4 2022 lúc 22:57
Bài 15: Chứng minh rằng các phân số sau là tối giản(n∈ N*)
a) dfrac{n+1}{2n+3} . b) dfrac{2n+3}{4n+8} .
c) dfrac{3n+1}{4n+1} .
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Bài 15: Chứng minh rằng các phân số sau là tối giản(n∈ N*)
a) \(\dfrac{n+1}{2n+3}\) . b) \(\dfrac{2n+3}{4n+8}\) .
c) \(\dfrac{3n+1}{4n+1}\) .
Lời giải:
a/
Gọi ƯCLN(n+1, 2n+3)=d$
Khi đó:
$n+1\vdots d\Rightarrow 2n+2\vdots d(1)$
$2n+3\vdots d(2)$
Từ $(1); (2)\Rightarrow (2n+3)-(2n+1)\vdots d$ hay $1\vdots d$
$\Rightarrow d=1$
Vậy $n+1, 2n+3$ nguyên tố cùng nhau nên phân số đã cho tối giản.
Câu b,c làm tương tự.
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Chứng minh rằng:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}=\dfrac{n+1}{2n+3}\)
Giúp mik vs các bn
\(P=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\\ 2P=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{\left(2n+1\right)\left(2n+3\right)}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n+1}-\dfrac{1}{2n+3}\\ =1-\dfrac{1}{2n+3}\\ =\dfrac{2\left(n+1\right)}{2n+3}\\ P=\dfrac{2\left(n+1\right)}{2n+3}:2\\ =\dfrac{n+1}{2n+3}\)
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28 tháng 3 2021 lúc 10:19
Chứng minh rằng :
a) \(\dfrac{1.3.5.....39}{21.22.23.....40}=\dfrac{1}{2^{20}}\)
b) \(\dfrac{1.3.5....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\dfrac{1}{2^n}\) với \(n\in\) N*
a) Vế trái \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)
\(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)
b) Vế trái
\(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)
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Tính các tổng sau:
Adfrac{1}{2.5}+dfrac{1}{5.8}+dfrac{1}{8.11}+.....+dfrac{1}{left(3n-1right)left(3n+2right)}
Bdfrac{1}{1.3.5}+dfrac{1}{3.5.7}+dfrac{1}{5.7.9}+....+dfrac{1}{left(2n-1right)left(2n+1right)left(2n+3right)}
Csqrt{1+dfrac{1}{1^2}+dfrac{1}{2^2}}+sqrt{1+dfrac{1}{2^2}+dfrac{1}{3^2}}+sqrt{1+dfrac{1}{3^2}+dfrac{1}{4^2}}+....+sqrt{1+dfrac{1}{2018^2}+dfrac{1}{2019^2}}
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Tính các tổng sau:
A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
B=\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}\)
C=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+....+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
\(\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}=\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(\Rightarrow C=2019-\dfrac{1}{2019}\)
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