Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Theo tính chất của dãy tỉ số bằng nha, ta có :
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=.....=\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\Rightarrow\dfrac{a_1}{a_2}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\dfrac{a_2}{a_3}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
.................................
\(\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
\(\Rightarrow\left(\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\right)^n=\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}........\dfrac{a_n}{a_{n+1}}\)
Vậy \(\left(\dfrac{a_1+a_2+......+a_n}{a_2+a_3+......+a_{n+1}}\right)=\dfrac{a_1}{a_{n+1}}\) (đpcm)
~ Học tốt ~
\(\dfrac{x_1}{a_1}=\dfrac{x_2}{a_2}=...=\dfrac{x_n}{a_n}=\dfrac{x_1+x_2+...+x_{n-1}+x_n}{a_1+a_2+...+a_{n-1}+a_n}\)
\(=\dfrac{c}{a_1+a_2+...+a_n}\)
Suy ra:
\(x_1=\dfrac{a_1.c}{a_1+a_2+...+a_n}\)
\(x_2=\dfrac{a_2.c}{a_1+a_2+...+a_n}\)
.........................................
\(x_n=\dfrac{a_n.c}{a_1+a_2+...+a_n}\)
áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=.....=\frac{an}{an+1}=\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\)
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\frac{a3}{a4}\cdot...\cdot\frac{an}{an+1}=\frac{a1}{an+1}=\left(\frac{a1}{a2}\right)^n=\left(\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\right)^n\)(vì từ 1 đến n có n chữ số)
=> đpcm
Bài 1:
a) \(\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)......\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\)
= \(\dfrac{-8}{9}.\dfrac{-9}{10}.......\dfrac{-2003}{2004}.\dfrac{-2004}{2005}\) = \(\dfrac{-8}{2005}\)
b) \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\) = \(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\)
= \(-2+\dfrac{1}{-2+\dfrac{1}{-1}}\) = \(-2+\dfrac{1}{-3}\) = \(\dfrac{-7}{3}\)
\(\text{Câu 1 : }\) Tính
\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)
\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=....=\left(\frac{a_n}{a_{n+1}}\right)^n=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)(1)
Ta có: \(\left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}....\frac{a_n}{a_{n+1}}=\frac{a_1}{a_{n+1}}\)(2)
Từ (1), (2) \(\Rightarrow\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)(đpcm)
\(\text{Áp dụng tính chất của dãy tỉ số bằng nhau có:}\)
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=...=\left(\frac{a_n}{a_{n+1}}\right)^n\)\(=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)
Mà\( \left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}\cdot\frac{a_1}{a_2}\cdot...\cdot\frac{a_1}{a_2}\)\(=\frac{a_1}{a_2}\cdot\frac{a_2}{a_3}\cdot...\cdot\frac{a_n}{a_{n+1}}\)\(=\frac{a_1}{a_{n-1}}\)
\(\Rightarrow\)\(\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)\(=\frac{a_1}{a_{n-1}}\)
bạn gõ phân số kiểu gì vậy
bn cho phân số z làm sao mà giải