Question

[haquynhanh]

tìm x biết :\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)

Answer 1

[Hoàng Ninh]

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)

\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)

\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)

\(\Rightarrow x=50\)

Vậy x = 50

Answer 2

[Emma Granger]

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)

\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)

\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)