Tìm x biết :
1/2 + 1/6 + 1/12 + ... + 1/x = 99/100
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LT
2
29 tháng 7 2017
1/1x2+1/2x3+1/3x4+...+1/x=99?100
1/1-1/2+1/2-1/3+1/3-1/4+...+1/x=99/100
1/1-1/x=99/100
1/x=1/1-99/100
1/x=1/100
=>x=100
kbn nha
BT
17 tháng 1 2018
ta có
1+2+3+.........+x=5050
=>\(\frac{x.\left(x+1\right)}{2}=5050\)
=>x.(x+1)=5050.2
=>x.(x+1)=10100
=>x.(x+1)=100.101
=>x=100
OD
1
8 tháng 4 2021
Giải:
1/2 + 1/6 + 1/12 + ... + 1/x.(x+1) = 99/100
=>1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x.(x+1) = 99/100
1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/x+1 = 99/100
1/1 - 1/x+1 = 99/100
1/x+1 = 1/1 - 99/100
1/x+1 = 1/100
=>x+1 = 100
x = 100 - 1
x = 99
Vậy ...
Chúc bạn học tốt!
OD
1
TA
8 tháng 4 2021
`1/2+1/6+1/12+....+1/(x(x+1))=99/100`
`-> 1/(1.2)+1/(2.3)+1/(3.4)+....+1/(x(x+1))=99/100`
`-> 1 - 1/2 + 1/2 -1/3 +1/3-1/4+....+1/x -1/(x+1)=99/100`
`-> 1-1/(x+1)=99/100`
`-> x/(x+1)=99/(99+1)`
`-> x=99`
OD
1
8 tháng 4 2021
Bài 1:
ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{x+1}=\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{x+1-1}{x+1}=\dfrac{99}{100}\)
\(\Leftrightarrow100x=99\left(x+1\right)\)
\(\Leftrightarrow100x-99x=99\)
hay x=99(thỏa ĐK)
Vậy: x=99
OD
1
8 tháng 4 2021
\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{99}{100}\)
\(\Leftrightarrow\)\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)=\(\dfrac{99}{100}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}\)=\(\dfrac{99}{100}\)
\(\Leftrightarrow\dfrac{1}{x+1}\)=\(\dfrac{1}{100}\)
\(\Leftrightarrow x+1\)=100
\(\Leftrightarrow\)\(x\)=99
DV
0
5 tháng 4 2017
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{99}{100}\)
Đặt \(x=n.\left(n+1\right)\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{n+1-n}{n.\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(=1-\frac{1}{\left(n+1\right)}=\frac{99}{100}\)
\(\frac{1}{\left(n+1\right)}=1-\frac{99}{100}=\frac{1}{100}\)
\(\Rightarrow x=\left(100-1\right).100\)
\(=9900\)
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}=\frac{99}{100}\)
Đề kì z