Ta có: \(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\) (1)
Xét vế trái ta có:
\(\left(\frac{10}{1.2}+\frac{10}{2.3}+...+\frac{10}{49.50}\right)+2x\)
\(=10.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(=10.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)+2x\)
\(=10.\left(1-\frac{1}{50}\right)+2x\)
\(=10.\frac{49}{50}+2x\)
\(=\frac{49}{5}+2x\) (2)
Xét vế phải ta có:
\(\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{47.49}-7x\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)-7x\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)-7x\)
\(=2.\left(1-\frac{1}{49}\right)-7x\)
\(=2.\frac{48}{49}-7x\)
\(=\frac{96}{49}-7x\) (3)
Từ (1), (2) và (3) => \(\frac{49}{5}+2x=\frac{96}{49}-7x\)
\(\Rightarrow2x+7x=\frac{96}{49}-\frac{49}{5}\)
\(\Rightarrow9x=\frac{480}{245}-\frac{2401}{245}\)
\(\Rightarrow9x=-\frac{1921}{245}\)
\(\Rightarrow x=-\frac{1921}{245}:9=-\frac{1921}{2205}\)
Vậy \(x=-\frac{1921}{2205}\)
Chúc bạn học tốt!
Ta có:\(\left(10-\frac{10}{2}+\frac{10}{2}-\frac{10}{3}+...+\frac{10}{49}-\frac{10}{50}\right)+2x=\left(2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+...+\frac{2}{47}-\frac{2}{49}\right)-7x\)
\(\left(10-\frac{10}{50}\right)+2x=\left(2-\frac{2}{49}\right)-7x\)
\(\frac{49}{5}+2x=\frac{96}{49}-7x\)
\(7x+2x=\frac{96}{49}-\frac{49}{5}\)
\(9x=-\frac{1921}{245}\)
\(x=-\frac{1921}{245}:9\)
\(x=-\frac{1921}{2205}\)
Vậy \(x=-\frac{1921}{2205}\)
