\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)

\(\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}.\frac{1}{2}\)

\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)

\(\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}=\frac{1}{9}\)

=>x+1=9

=>x=8

Lưu ý : . là dấu nhân

ai làm đúng , nhanh mình k

bình chọn đúng

co \(\frac{1}{9\cdot10}=\frac{1}{9}-\frac{1}{10}\)

\(\frac{1}{10\cdot11}=\frac{1}{10}-\frac{1}{11}\)

............

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

nen \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\)

\(=\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}-...+\frac{1}{x}-\frac{1}{x+1}\)

=\(\frac{1}{9}-\frac{1}{x+1}\)

2 .  ( \(\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+...+\frac{1}{x\left(x+1\right)}\))

=  2  . ( \(\frac{1}{9}-\frac{1}{x+1}\))  = \(\frac{2}{9}-\frac{2}{x+1}\)

Bạn ơi tim x mà bạn

bo may ko biet

=1/10 - 1/11+......+1/99-1/100

=1/10-1/100

=9/100

Ta có A=1/10.11+1/11.12+...+1/98.99+1/99.100

            =1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100

            =1/10-1/100

           =10/100-1/100

          =9/100

Vậy A=9/100

Giải:

A=1/10.11+1/11.12+...+1/98.99+1/99.100

A=1/10-1/11+1/11-1/12+...+1/98-1/99+1/99-1/100

A=1/10-1/100

A=9/100

Chúc bạn học tốt!

\(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}\)

\(=\frac{8-7}{7\cdot8}+\frac{9-8}{8\cdot9}+\frac{10-9}{9\cdot10}+\frac{11-10}{10\cdot11}+\frac{12-11}{11\cdot12}+\frac{13-12}{12\cdot13}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)

\(=\frac{1}{7}-\frac{1}{13}=\frac{13-7}{7\cdot13}=\frac{6}{91}\)

\(\frac{6}{91}\)nha

1) A=7/10.11+7/11.12+7/12.13+...+7/69.70

  A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)

 A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)

 A= 7.(1/10-1/70)

 A=7.3/35=3/5

2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75

  B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

  B=1/25-1/75=2/75

A = 7/ 10.11 + 7/ 11.12 + 7/ 12.13 + .... + 7/69.70

(1/7).A=1/10.11+1/11.12+...+1/69.70

=1/10-1/11+1/11-1/12+...+1/69-1/70

=1/10-1/70=3/35

=>A=7.(3/35)

=3/5

2 ) B = 1/ 25.27 + 1/ 27.29 + 1/29.31+ ......+ 1/ 73.75

=>(1/2).B=2/25.27+...+2.73.75

=1/25-1/27+...+1/73-1/75

=1/25-1/75

=2/75

=>B=4/75

 A=7.(1/10.11+1/11.12+1/12.13+...+1/69.70)

 A= 7.(1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70)

 A= 7.(1/10-1/70)

 A=7.3/35=3/5

2)B=1/25.27+1/27.29+1/29.31+...+1/73.1/75

  B=1/25-1/27+1/27-1/29+1/29-1/31+...+1/73-1/75

  B=1/25-1/75=2/75

\(2\left(\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{9}\)

\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1}{9}\)

\(\Leftrightarrow2\left(\frac{1}{9}-\frac{1}{x+1}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{9}\div2\)

\(\Leftrightarrow\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{9}-\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=18-1\)

\(\Leftrightarrow x=17\)

\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\frac{5}{3}+\frac{3}{4}\)

\(\Leftrightarrow\left|x\right|=\frac{20}{12}+\frac{9}{12}\)

\(\Leftrightarrow\left|x\right|=\frac{29}{12}\)

\(\Leftrightarrow x=\pm\frac{29}{12}\)

Bài 1 :

\(2.\left(\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{x.\left(x+1\right)}\right)\)= \(\frac{1}{9}\)

=> \(\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\)= \(\frac{1}{18}\)

\(\frac{1}{9}-\frac{1}{x+1}=\frac{1}{18}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

=> x = 17

vậy x = 17

đặt a=1/10.11+ 1/11.12+..+1/99.100

        =1/10-1/11+1/11-1/12+...+1/99-1/100

        =1/10-1/100=9/100

    vậy a=9/100

9/100 lấy 1/10-1/100 là ra

9/100