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\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{996}{997}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{996}{997}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{996}{997}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{997}\)
\(\Rightarrow x+1=997\)
\(\Rightarrow x=996\)
\(\Leftrightarrow\)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
\(\Leftrightarrow\)\(\frac{x}{x+1}\)\(=\frac{996}{997}\)
\(\Leftrightarrow x=996\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{y\times\left(y+1\right)}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{y}-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow1-\frac{1}{y+1}=\frac{996}{997}\)
\(\Leftrightarrow\frac{1}{y+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Leftrightarrow y+1=997\Leftrightarrow y=996\)
Vậy y = 996
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{996}{997}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{996}{997}\)
\(1-\frac{1}{x+1}=\frac{996}{997}\)
\(\frac{1}{x+1}=1-\frac{996}{997}\)
\(\frac{1}{x+1}=\frac{1}{997}\)
\(\Rightarrow x+1=997\)
\(x=997-1\)
\(x=996\)
1/1*2 + 1/2*3 + 1/3*4 + .... + 1/99 * 100
= 1- 1/100
= 99/100
Ta có : A = \(\frac{1}{1\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{3\text{x}4}+...+\frac{1}{X\text{x}\left(X+1\right)}\)
A = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
A = \(\frac{1}{1}-\frac{1}{x+1}\)
A = \(\frac{x}{x+1}\)
Ủng hộ mik nhá !!!!
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=?\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{1}-?\)
\(\Rightarrow x+1=?\Leftrightarrow x=?\)
Đặt A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow A=1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\Rightarrow\)\(\frac{1}{x+1}=\frac{1}{2014}\)
\(\Rightarrow x+1=2014\)
\(\Rightarrow x=2014-1\)
\(\Rightarrow x=2013\)
Vậy x=2013
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2014}\)
\(1-\frac{1}{x+1}=\frac{2013}{2014}\)
\(\frac{1}{x+1}=1-\frac{2013}{2014}\)
\(\frac{1}{x+1}=\frac{1}{2014}\)
Vì \(x+1\)là mẫu số nên:
\(x+1=2014\)
\(x=2014-1=2013\)
Vậy ....
P/s: Dấu . là nhân nha!
`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`
`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`
`=>x/(x+1)=1-1/32`
`=>x/(x+1)=31/32`
`=>32x=31(x+1)`
`=>32x=31x+31`
`=>32x-31x=31`
`=>x=31`
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{996}{997}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{996}{997}\)
\(1-\frac{1}{x+1}=\frac{996}{997}\)
\(\frac{1}{x+1}=1-\frac{996}{997}=\frac{1}{997}\)
\(\Rightarrow x+1=997\Rightarrow x=996\)
Vậy \(x=996\)