Được chưa bây giờ tic đúng đi

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

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

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

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

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

\(\Rightarrow x+1=997\)

               \(x=997-1\)

               \(x=996\)

Ấm đầu hả phải tic bài làm chứ 

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

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

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

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

\(\Rightarrow x+1=997\)

\(\Rightarrow x=996\)

\(\Leftrightarrow\)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

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

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 

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

cảm ơn

mình làm luộn 2 câu còn lại nhé ^^

.c,\(|x|-\frac{15}{2}=\frac{15}{4}\)

.\(< =>|x|=\frac{15}{4}+\frac{15}{2}=\frac{15}{4}+\frac{30}{4}=\frac{45}{4}\)

.\(< =>\orbr{\begin{cases}x=\frac{45}{4}\\x=-\frac{45}{4}\end{cases}}\)

.d,\(|\frac{3}{4}-x|+1=\frac{3}{2}\)

.\(< =>|\frac{3}{4}-x|=\frac{3}{2}-1=\frac{3}{2}-\frac{2}{2}=\frac{1}{2}\)

.\(< =>\orbr{\begin{cases}\frac{3}{4}-x=\frac{1}{2}\\\frac{3}{4}-x=-\frac{1}{2}\end{cases}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\\x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\end{cases}}}\)

1. B = 1+ (2+ 3 +4+.... +98 +99)                  

        = 1+ 98

         = 99

2