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 ~

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

\(\text{a) }\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{10}-1\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-1\right)\\ =\left(1-\dfrac{9}{9}\right)\left(\dfrac{1}{10}-\dfrac{10}{10}\right)...\left(\dfrac{1}{2004}-1\right)\left(\dfrac{1}{2005}-\dfrac{2005}{2005}\right)\\ =\dfrac{-8}{9}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-2003}{2004}\cdot\dfrac{-2004}{2005}\\ =\dfrac{\left(-8\right)\cdot\left(-9\right)\cdot..\cdot\left(-2003\right)\cdot\left(-2004\right)}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8\cdot9\cdot...\cdot2003\cdot2004}{9\cdot10\cdot...\cdot2004\cdot2005}\\ =-\dfrac{8}{2005}\)

\(-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{-2+3}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-2+\dfrac{1}{1}}}\\ =-2+\dfrac{1}{-2+\dfrac{1}{-1}}\\ =-2+\dfrac{1}{-3}\\ =-2+\dfrac{-1}{3}=-\dfrac{7}{3}\)

Ta có :

\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)

\(\Rightarrow\)\(\frac{a_1^n}{a_2^n}=\frac{a_2^n}{a_3^n}=...=\frac{a_n^n}{a_{n+1}^n}=\frac{a_1^n+a_2^n+...+a_n^n}{a_2^n+a_3^n+...+a_{n+1}^n}=\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}=\frac{a_1.a_2...a_n}{a_2.a_3...a_{n+1}}=\frac{a_1}{a_{n+1}}\)

\(\dfrac{x_1}{a_1}=\dfrac{x_2}{a_2}=...=\dfrac{x_n}{a_n}=\dfrac{x_1+x_2+...+x_{n-1}+x_n}{a_1+a_2+...+a_{n-1}+a_n}\)

\(=\dfrac{c}{a_1+a_2+...+a_n}\)

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 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\( \left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}\cdot\frac{a_1}{a_2}\cdot...\cdot\frac{a_1}{a_2}\)\(=\frac{a_1}{a_2}\cdot\frac{a_2}{a_3}\cdot...\cdot\frac{a_n}{a_{n+1}}\)\(=\frac{a_1}{a_{n-1}}\)

\(\Rightarrow\)\(\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)\(=\frac{a_1}{a_{n-1}}\)