Question

Tìm x : 4/1x3 + 4/3x5 + ... + 4/99x101 -x - 200/101 = 1

Answer 1

\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+...+\frac{4}{99\cdot101}-x-\frac{200}{101}=1\)

\(\frac{4}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)-x=1+\frac{200}{101}\)

\(\frac{4}{2}\cdot\left(1-\frac{1}{101}\right)-x=\frac{301}{101}\)

\(\frac{4}{2}\cdot\frac{100}{101}-x=\frac{301}{101}\)

\(\frac{200}{101}-x=\frac{301}{101}\)

\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)

Answer 2

Ta có : \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}-x-\frac{200}{101}=1\)

\(\Rightarrow\)\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=1+\frac{200}{101}+x\)

=> \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+.....+\frac{4}{99.101}=\frac{301}{101}+x\)

=> \(2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\right)=\frac{301}{101}+x\)

=> \(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)=\frac{301}{101}+x\)

=> \(2\left(1-\frac{1}{101}\right)=\frac{301}{101}+x\)

=> \(2.\frac{100}{101}=\frac{301}{101}+x\)

=> \(\frac{200}{101}=\frac{301}{101}+x\)

\(\Rightarrow x=\frac{301}{101}-\frac{200}{101}=1\)