tìm x
1/2.3+1/3.4+1/4.5+......+1/x.(x+1)=299/60000
BT2: Tìm x\(\in\) N*, biết
2) \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{x.\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{300}{600}-\dfrac{299}{600}\)
\(\dfrac{1}{x+1}=\dfrac{1}{600}\)
=> x + 1 = 600
x = 600 - 1
x = 599
Vậy x = 599
tìm x
1/2.1+1/2.3+1/3.4+1/4.5+.....+1/x(x+1)
Tìm một số tự nhiên x biết: 1/2.3 + 1/3.4 + 1/4.5 +....+ 1/x.(x+1)=9/20
(chấm là nhân)
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(x+1=20\)
\(x=20-1\)
\(x=19\)
Tìm một số tự nhiên x biết: 1/2.3 + 1/3.4 + 1/4.5 +....+ 1/x.(x+1)=9/20
(chấm là nhân)
Có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(\Rightarrow x+1=20\Leftrightarrow x=19\)
Tìm X biết 2/2.3 + 2/3.4 + 2/4.5+...... +2/X(X+1) = 2007/2009
TÌM x THUỘC N:2/2.3+2/3.4+2/4.5+....+2/x.(x+1)=1999/2001
Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
\(\Rightarrow\)x+1=2001
x=2000
Vậy x=2000.
Tìm x khác 0, biết :\(1/2.3+1/3.4+1/4.5+1/5.6+...+1/19.20=18/x\)
.3-2/2.3 + 4-3/3.4 + 5-4/4.5 + 6-5/5.6 +...+ 20-19/19.20=18/x
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +...+ 1/19 - 1/20=18/x
1/2 - 1/20=18/x
10/20 - 1/20=18/x
9/20=18/x
18/40=18/x
=>x=40
Vậy x=40
tìm x:1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=16/51
1+2.(1/2.3+1/3.4+1/4.5+...+1/x.(x+1)=3/2
1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2
1+2.(1/2-1/x+1)=3/2
1-2/x+1=3/2-1
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