Question

(1/1.2.3+1/2.3.4+1/3.4.5+1/4.5.6+1/5.6.7+...+1/8.9.10). x=23/45

các bạn giúp mình nhé , giải một cách thật nhanh mình mới tk

Answer 1

Ta có: \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

Answer 2

Đặt A=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)

    2A=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\)

    2A=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}\) \(+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

    2A=\(\frac{1}{1.2}-\frac{1}{9.10}\)

    2A=\(\frac{22}{45}\)

      A=\(\frac{22}{45}\div2\)

      A=\(\frac{11}{45}\)

\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)

             \(x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)

Vậy x=\(\frac{23}{11}\)