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Phần chứng tỏ quy đồng lên rồi tính là ra
Còn phần tính S thì áp dụng tính chất vừa chứng tỏ để tách ra
Kết quả là 49/50
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Ta có: 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2/3
=> 1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2/3
=>1-1/x+1=2/3
=>1/x+1=1/3
=>3=x+1
=>x=2
Ta có\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)
=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)
=>\(1-\frac{1}{x+1}=\frac{2}{3}\)
=>\(\frac{1}{x+1}=1-\frac{2}{3}\)
=>\(\frac{1}{x+1}=\frac{1}{3}\)
=>\(x+1=3\)
=>\(x=2\)
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(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)
(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1 - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))
\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = \(\dfrac{1}{2003}\)
\(x\) = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)
\(x\) = \(\dfrac{2022}{2003}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có 1+(1+2)+(1+2+3)+...+(1+2+3+...+100)
=4+(1+3).3/2+9+(1+4).4/2+...+(1+100).100/2
=1/2(1.2+2.3+.....+100.101)
=>1/2.100.101.102
con cái dưới thì bằng 99.100.101
=>F=51/99
ngu rua mà ko biet lam
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{4}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{99.100}\)
\(=\frac{3}{4}+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{3}{4}+1-\frac{1}{100}=\frac{87}{50}\)
(1-1/2).(1-1/3).(1-1/4)...(1-1/2002).x=1-1/1.2-1/2.3-1/3.4-...-1/2002.2003 ghi loi giai nha ae