Nguyễn Thị Mai:
\(^{\frac{1}{2.3}x+\frac{1}{3.4}x+....+\frac{1}{49.50}x=1}\)
Đặt: \(^{\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}+0+0+...+0-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\left(\frac{1}{2}-\frac{1}{50}\right)x=1}\)
\(^{\Leftrightarrow\frac{12}{25}x=1}\)
\(^{\Leftrightarrow x=\frac{25}{12}}\)
Tham khảo nha ~~
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