Câu 1:
* \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+2y-3z}{5+8-9}=\dfrac{x+2y-3z}{4}\) (1)
* \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x-2y+3z}{5-8+9}=\dfrac{x-2y+3z}{6}\) (2)
(1)(2) => \(\dfrac{x+2y-3z}{4}=\dfrac{x-2y+3z}{6}=\dfrac{x+2y-3z}{x-2y+3z}=\dfrac{4}{6}=\dfrac{2}{3}.\)
=> P = \(\dfrac{2}{3}\)
Câu 7:
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ \Leftrightarrow3A-A=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\\ \Leftrightarrow2A=1-\dfrac{1}{3^{99}}\\ \Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{3^{99}\cdot2}< \dfrac{1}{2}\left(\dfrac{1}{3^{99}\cdot2}>0\right)\)
