\(x-\dfrac{2}{3}=2x+\dfrac{1}{5}\\ 2x-x=-\dfrac{1}{5}-\dfrac{2}{3}\\ x=-\dfrac{13}{15}\)

\(\dfrac{-6}{12}=\dfrac{x}{8}=\dfrac{-7}{y}=\dfrac{z}{-18}\\ \Leftrightarrow\dfrac{-1}{2}=\dfrac{x}{8}=\dfrac{-7}{y}=\dfrac{z}{-18}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{8\cdot\left(-1\right)}{2}=-4\\y=\dfrac{\left(-7\right)\cdot2}{-1}=14\\z=\dfrac{\left(-18\right)\cdot\left(-1\right)}{2}=9\end{matrix}\right.\)

vậy x = -4; y = 14; z = 9

quả bóng sẽ không đạt tới độ cao ban đầu

vì khi nảy lên lại, quả bóng sẽ cọ xát với không khí, một phần năng lượng ban đầu sẽ chuyển hoá thành nhiệt năng vì quả bóng cọ xát với không khí

\(\dfrac{15}{x}=\dfrac{y}{4}=\dfrac{z}{16}=\dfrac{6}{-8}\\ \Leftrightarrow\dfrac{15}{x}=\dfrac{y}{4}=\dfrac{z}{16}=\dfrac{-3}{4}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{4\cdot15}{-3}=-20\\y=\dfrac{\left(-3\right)\cdot4}{4}=-3\\z=\dfrac{\left(-3\right)\cdot16}{4}=-12\end{matrix}\right.\\ \text{vậy }x=-20;y=-3;z=-12\)

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{4950}\\ A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{9900}\\ A=2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\\ A=2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\right)\\ A=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ A=2\cdot\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\\ A=2\cdot\dfrac{49}{100}\\ A=\dfrac{49}{50}\)

bạn bị lộn ngay chỗ \(9\cdot\dfrac{3}{80}=\dfrac{21}{80}\)

dòng thứ 6; 7

\(A=\dfrac{3^2}{20\cdot23}+\dfrac{3^2}{23\cdot26}+...+\dfrac{3^2}{77\cdot80}\\ A=\dfrac{3^2}{3}\cdot\left[\left(\dfrac{1}{20}-\dfrac{1}{23}\right)+\left(\dfrac{1}{23}-\dfrac{1}{26}\right)+...+\left(\dfrac{1}{77}-\dfrac{1}{80}\right)\right]\\ A=3\cdot\left[\dfrac{1}{20}-\dfrac{1}{80}\right]\\ A=3\cdot\dfrac{3}{80}=\dfrac{9}{80}< 1\)