Question

Bài 1:Tìm x 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{215}{216}\)

giúp mình cách trình bày với nhé mai phải nộp rồi

Answer 1

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)

\(\Leftrightarrow x=216-1=215\)