\(x+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{9900}\right)=2\)

=> \(x+\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{99.100}\right)=2\)

 => \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2\)(100 hạng tử x)

=> \(100x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=2\)

=>  \(100x+1-\frac{1}{100}=2\)

=> \(100x+\frac{99}{100}=2\)

=> \(100x=\frac{101}{100}\)

=> \(x=\frac{101}{10000}\)

\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)...\left(1-\frac{2}{9900}\right)\)

\(=\frac{4}{6}.\frac{10}{12}...\frac{9898}{9900}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{98.101}{99.100}\)

\(=\frac{1.2...98}{3.4...100}.\frac{4.5...101}{2.3...99}\)

\(=\frac{2}{99.100}.\frac{100.101}{2.3}\)

\(=\frac{101}{99.3}\)

\(=\frac{101}{297}\)

đáp số:\(\frac{101}{297}\)

ai k mk mk sẽ k lại ^-^

là 101/247

l

\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{9900}=7,5\)

\(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=7,5\)

\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=7,5\)

\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)=7,5\)

\(x-\left(1-\frac{1}{100}\right)=7,5\)

\(x-\frac{99}{100}=7,5\)

\(\Rightarrow x=7,5+\frac{99}{100}=\frac{750}{100}+\frac{99}{100}\)

\(\Rightarrow x=\frac{849}{100}=8,49\)

Dấu "." là dấu nhân nha bạn

tớ chưa hiểu lắm

Đặt A=1+2+2²+2³+...+2¹⁰⁰

suy ra 2A=2+2²+2³+...+2¹⁰⁰

suy ra 2A-A=(2+2²+2³+...+2¹⁰¹)-(1+2+2²+2³+...+2¹⁰⁰⁾

                 =2¹⁰¹-1

Vậy 1+2+2²+2³+...+2¹⁰⁰=2¹⁰¹-1

 A=1+2+2^2+2^3+...+2^100

2A=2+2²+2³+2⁴+...+2¹⁰¹

2A-A=2¹⁰¹-1

Vậy A= 2¹⁰¹-1

1) x=100

    1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100

= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101

= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101

= 1/1 - 1/101

= 100 /101

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

=\(1-\frac{1}{101}\)

=\(\frac{100}{101}\)

Mik trả lời ở bài dưới rồi đó.

o dau ha ho thu giang

Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)

     \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

     \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

     \(=1-\frac{1}{101}\)

     \(=\frac{100}{101}\)

Tương đương \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)