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:

\(x_1=\dfrac{a_1.c}{a_1+a_2+...+a_n}\)

\(x_2=\dfrac{a_2.c}{a_1+a_2+...+a_n}\)

.........................................

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

Sai đề.

5a

Ta có \(\dfrac{a}{b}=\dfrac{a^2}{b^2}\) ; \(\dfrac{c}{d}=\dfrac{c^2}{d^2}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\)=> \(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=>\(\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\)=\(\dfrac{a^2+c^2}{b^2+d^2}\)(T/c cuả dãy tỉ số bằng nhau)

=> ĐPCM

Bài 1:

Ta có: \(\dfrac{U+2}{U-2}=\dfrac{V+3}{V-3}\)

\(\Rightarrow\left(U+2\right)\left(V-3\right)=\left(U-2\right)\left(V+3\right)\)

\(\Rightarrow\left(U+2\right)V-\left(U+2\right)3=\left(U-2\right)V+\left(U-2\right)3\)

\(\Rightarrow UV+2V-3U-6=UV-2V+3U-6\)

\(\Rightarrow2V-3U=3U-2V\)

\(\Rightarrow2U+2U=3V+3V\)

\(\Rightarrow2\times2U=2\times3V\)

\(\Rightarrow2U=3V\)

\(\Rightarrow\dfrac{U}{3}=\dfrac{V}{2}\)

(Như mình đã nói ở Câu hỏi của ARMY, cách này tuy dài nhg chặt hơn)

(Tiếp tục nè)

\(\Rightarrow\dfrac{U^2}{9}=\dfrac{V^2}{4}=\dfrac{U^2+V^2}{9+4}=\dfrac{52}{13}=4\)

\(\Rightarrow U^2=4\times9=36\Rightarrow U=\pm6\)

\(V^2=4\times4=16\Rightarrow V=\pm4\)

Vì \(\dfrac{U}{3}=\dfrac{V}{2}\Rightarrow U,V\) cùng dấu

\(\Rightarrow\left(U,V\right)=\left(6;4\right)\) và \(\left(U,V\right)=\left(-6;-4\right)\)

Vậy...

Bài 2:

Vì:\(\dfrac{x}{y}=\dfrac{z}{t}\Rightarrow\dfrac{x}{z}=\dfrac{y}{t}=\dfrac{x+y}{z+t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\left(\dfrac{x+y}{z+t}\right)^2\left(1\right)\)

Ta còn có: \(\dfrac{x}{z}=\dfrac{y}{t}\Rightarrow\left(\dfrac{x}{z}\right)^2=\left(\dfrac{y}{t}\right)^2=\dfrac{xy}{zt}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow\).....

Chúc bn học tốt

P/S: Những chỗ mình ghi chấm chấm có nghĩa là bn tự kết nốt