1) Chung minh cong thuc ghep noi tiep:
\(\dfrac{1}{C_{td}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+...+\dfrac{1}{C_n}\)
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Những câu hỏi liên quan
10 tháng 1 2019
Ta có:
\(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}=1\Rightarrow\left(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}\right)^2=1\)
\(\Rightarrow\dfrac{a^2_2}{a^2_1}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}+2\left(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{c_2a_2}{a_1c_1}\right)=1\)
\(\Rightarrow\dfrac{a_2^2}{a^2_1}+\dfrac{b^2_2}{b^2_1}+\dfrac{c^2_2}{c^2_1}+2\left(\dfrac{a_2b_2c_1+b_2c_2a_1+c_2a_2b_1}{a_1b_1c_1}\right)=1\)(1)
Theo giả thiết:
\(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\Leftrightarrow\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_2b_2c_2}=0\)(2)
Từ (1) và (2) suy ra đpcm
10 tháng 1 2019
Đặt \(\dfrac{a_1}{a_2}=p;\dfrac{b_1}{b_2}=q;\dfrac{c_1}{c_2}=r\), có:
\(p+q+r=0\) (1)
\(\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=1\) (2)
Từ (2) => \(\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}+2\dfrac{p+q+r}{pqr}=1\)
Kết hợp với (1), ta được: \(\dfrac{1}{p^2}+\dfrac{1}{q^2}+\dfrac{1}{r^2}=1\Rightarrow\dfrac{a^2_2}{a^2_1}+\dfrac{b^2_2}{b_1^2}+\dfrac{c_2^2}{c^2_1}=1\left(đpcm\right)\)
27 tháng 2 2017
1. Ta có \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\left(b+c\right)\left(\dfrac{a}{b+c}\right)+\dfrac{b^2}{c+a}+\left(c+a\right)\left(\dfrac{b}{c+a}\right)+\dfrac{c^2}{a+b}+\left(a+b\right)\left(\dfrac{c}{a+b}\right)=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\) (đpcm).
27 tháng 2 2017
2. Ta có: \(\dfrac{a_1}{a_2}+\dfrac{b_1}{b_2}+\dfrac{c_1}{c_2}=0\)
\(\Rightarrow\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_2b_2c_2}=0\)
\(\Rightarrow a_1b_2c_2+b_1a_2c_2+c_1a_2b_2=0\)
Lại có: \(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}=1\)
\(\Rightarrow\left(\dfrac{a_2}{a_1}+\dfrac{b_2}{b_1}+\dfrac{c_2}{c_1}\right)^2=1\)
\(\Rightarrow\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}+2\left(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}\right)=1\)
Mặt khác: \(\dfrac{a_2b_2}{a_1b_1}+\dfrac{b_2c_2}{b_1c_1}+\dfrac{a_2c_2}{a_1c_1}=\dfrac{a_1b_2c_2+b_1a_2c_2+c_1a_2b_2}{a_1b_1c_1}=0\)
Vậy \(\dfrac{a_2^2}{a_1^2}+\dfrac{b_2^2}{b_1^2}+\dfrac{c_2^2}{c_1^2}=1\) (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 11 2021
Xét khai triển:
\(\left(1+x\right)^{2017}=C_{2017}^0+xC_{2017}^1+x^2C_{2017}^2+...+x^{2017}C_{2017}^{2017}\)
Lấy tích phân 2 vế:
\(\int\limits^1_0\left(1+x\right)^{2017}=\int\limits^1_0\left(C_{2017}^0+xC_{2017}^1+...+x^{2017}C_{2017}^{2017}\right)\)
\(\Leftrightarrow\dfrac{2^{2018}-1}{2018}=C_{2017}^0+\dfrac{1}{2}C_{2017}^1+...+\dfrac{1}{2018}C_{2017}^{2017}\)
Vậy \(S=\dfrac{2^{2018}-1}{2018}\)
18 tháng 7 2018
ta có : \(Q=C^1_n+2\dfrac{C_n^2}{C_n^1}+...+k\dfrac{C^k_n}{C_n^{k-1}}+...+n\dfrac{C^n_n}{C_n^{n-1}}\)
\(\Leftrightarrow Q=\dfrac{n!}{1!\left(n-1\right)!}+2\dfrac{1!\left(n-1\right)!}{2!\left(n-2\right)!}+...+k\dfrac{\left(k-1\right)!\left(n-k+1\right)!}{k!\left(n-k\right)!}+...+\dfrac{n\left(n-1\right)!1!}{n!}\)
\(\Leftrightarrow Q=n+\dfrac{2\left(n-1\right)}{2}+...+\dfrac{k\left(n-k+1\right)}{k}+...+\dfrac{n}{n}\)
\(\Leftrightarrow Q=n+\left(n-1\right)+...+\left(n-k+1\right)+...+1\)
\(\Leftrightarrow Q=n^2-\left(1+\left(1+1\right)+\left(1+2\right)+...+\left(n-1\right)\right)\)