Question

[Phạm Đức Huy]

\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+....+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{1}{3}\)-\(\dfrac{x+1}{324}\)
giúp mình vói ạ

Answer 1

[Minh Hiếu]

\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)

\(\left(x+1\right)^2=324=18^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=18\\x+1=-18\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-19\end{matrix}\right.\)

Answer 2

[Nguyễn Thùy Anh]

Ta có \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{x+1}{324}\)

\(\Rightarrow\)\(\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{x+1}{324}+\dfrac{1}{x+1}\)

\(\Rightarrow\)\(\dfrac{1}{x+1}-\dfrac{x+1}{324}=0\)

\(\Rightarrow\)\(\dfrac{1}{x+1}=\dfrac{x+1}{324}\)

\(\Rightarrow\)(x+1).(x+1)=324

\(\Rightarrow\)(x+1)²=324

\(\Rightarrow\)(x+1)² = 18² = (-18)²

TH1: (x+1)² = 18²

\(\Rightarrow\)x+1 = 18 

\(\Rightarrow\)x = 17

TH2: (x+1)² = (-18)²

\(\Rightarrow\)x+1 = -18 

\(\Rightarrow\)x = -19

Vậy x\(\in\)\(\left\{17;-19\right\}\)