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

49/50

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

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{3}\)

=>\(1-\frac{1}{x+1}=\frac{2}{3}\)

=>\(\frac{1}{x+1}=1-\frac{2}{3}\)

=>\(\frac{1}{x+1}=\frac{1}{3}\)

=>\(x+1=3\)

=>\(x=2\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{3}\)

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

\(\Rightarrow1-\frac{1}{x-1}=\frac{2}{3}\)

\(\Rightarrow\frac{1}{3}=\frac{1}{x-1}\)

\(\Rightarrow x=3+1=4\)

Lời giải :

Đặt S=1.2+2.3+3.4+4.5+…+99.100+100.101

3S=1.2.3+2.3.3+3.4.3+4.5.3+…+99.100.3+100.101.3

=1.2(3−0)+2.3(4−1)+3.4(5−2)+4.5(6−3)+…+99.100(101−98)+100.101(102−99)

=0.1.2-1.2.3+1.2.3-2.3.4+...+99.100.101-100.101.102

=100.101.102

S=100.101.34=343400

1.Tính 

a) Ta có: 

  A=(1-1/2²).(1-1/3²)...(1-1/100²)

=>A=3/2².8/3².....9999/100²

=>A=(1.3/2.2).(2.4/3.3).....(99.101/100.100)

=>A=(1.2.3.....99/2.3.4.....100).(3.4.5.....101/2.3.4.....100)

=>A=1/100.101/2

=>A=101/200

b) Ta có: 

  B=-1/1.2-1/2.3-1/3.4-...-1/100.101

=>B=-(1/1.2+1/2.3+1/3.4+...+1/100.101)

=>B=-(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)

=>B=-(1-1/101)

=>B=-100/101

 c) Ta có:

 C=1.2+2.3+3.4+...+100.101

       =>3C=1.2.3+2.3.3+3.4.3+...+100.101.3

       =>3C=1.2.3+2.3.(4-1)+3.4.(5-2)+...+100.101.(102-99)

       =>3C=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+...+100.101.102

       =>3C=100.101.102

       =>3C=1030200

       =>C=343400

Chúc bạn hok tốt nhé >:)!!!!!

(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 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))

   \(\dfrac{1}{2022}\).\(x\)        =    \(\dfrac{1}{2003}\)

             \(x\)        = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)

             \(x\)       =     \(\dfrac{2022}{2003}\)

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