\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{996}{997}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{996}{997}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{996}{997}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{997}\)

\(\Rightarrow x+1=997\)

\(\Rightarrow x=996\)

\(\Leftrightarrow\)1-1/2+1/2-1/3+1/3-1/4+..+1/x-1/(x+1)=996/997

\(\Leftrightarrow\)1-1/(x+1)=996/997

\(\Leftrightarrow\)\(\frac{x}{x+1}\)\(=\frac{996}{997}\)

\(\Leftrightarrow x=996\)

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{Xx\left(X+1\right)}=\frac{996}{997}\)

Xét dạng tổng quát : \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{ax\left(a+1\right)}-\frac{a}{ax\left(a+1\right)}=\frac{a+1-a}{ax\left(a+1\right)}=\frac{1}{ax\left(a+1\right)}\)

Do đó \(\frac{1}{ax\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)

Áp dụng \(\frac{1}{1x2}=\frac{1}{1x\left(1+1\right)}=\frac{1}{1}-\frac{1}{1+1}=\frac{1}{1}-\frac{1}{2}\)

\(\frac{1}{2x3}=\frac{1}{2x\left(2+1\right)}=\frac{1}{2}-\frac{1}{2+1}=\frac{1}{2}-\frac{1}{3}\)

.......

\(\frac{1}{Xx\left(X+1\right)}=\frac{1}{X}-\frac{1}{X+1}\)

Do đó \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{Xx\left(X+1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{X}-\frac{1}{X+1}\)

\(=\frac{1}{1}-\frac{1}{X+1}=1-\frac{1}{X+1}=\frac{X+1}{X+1}-\frac{1}{X+1}=\frac{X}{X+1}\)

Vì \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{Xx\left(X+1\right)}=\frac{996}{997}\)

nên \(\frac{X}{X+1}=\frac{996}{997}\)

\(\frac{X}{X+1}=\frac{996}{996+1}\)

Vậy X=996

Bài này đề bài là giải phương trình hở bạn :

Gỉai 

Phương trình đẫ cho trên đề bài tương đương với :

\(\frac{x+1}{1000}+1+\frac{x+2}{999}+1+\frac{x+3}{998}+1+\frac{x+4}{997}+1+\frac{x+5}{996}+1+\frac{x+6}{995}+1=0\)

\(\Leftrightarrow\frac{x+1001}{1000}+\frac{x+1001}{999}+\frac{x+1001}{998}+\frac{x+1001}{997}+\frac{x+1001}{996}+\frac{x+1001}{995}=0\)

\(\Leftrightarrow\left(x+1001\right)\left(\frac{1}{1000}+\frac{1}{999}+\frac{1}{998}+\frac{1}{997}+\frac{1}{996}+\frac{1}{995}\right)=0\)

\(\Leftrightarrow x=-1001\)

Vậy nghiêm của phương trình là : \(x=-1001\)

Chúc bạn học tốt !!!

hoang viet nhat

Làm thiếu giải thích 

đáp số là 996 đúng ko

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{996}{997}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)= \(\dfrac{996}{997}\) \(1-\dfrac{1}{x+1}\) = \(\dfrac{996}{997}\)

\(\dfrac{1}{x+1}\) = \(1-\dfrac{996}{997}\)

\(\dfrac{1}{x+1}\) =\(\dfrac{1}{997}\)

\(\Rightarrow\) x + 1 = 997

x = 997 - 1

x = 996

Vậy x = 996