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

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

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

  =\(\frac{1}{2.x}+\left(1-\frac{1}{100}\right)\)\(=2\)

  =\(\frac{1}{2.x}+\frac{99}{100}\)\(=2\)

  =\(\frac{1}{2.x}=2-\frac{99}{100}\)

  =\(\frac{1}{2.x}=\frac{101}{200}\)

  =\(2.x=200\)

  =\(x=200:2=100\)

1/2 * x + 1/2 + 1/6 + 1/12 + .... + 1/9900 = 2 

<=> 1/2 * x + ( 1/2 + 1/6 + 1/12 + ... + 1/9900 ) = 2 

<=> 1/2 * x + ( 1 /1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 ) = 2

<=> 1/2 * x + ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 ) = 2 

<=> 1/2 * x + ( 1 - 1/100 ) = 2 

<=> 1/2 * x + ( 100/100 - 1/100 ) = 2 

<=> 1/2 * x + 99/100 = 2 

<=> 1/2 * x = 2 - 99/100 

<=> 1/2 * x = 101/100

<=> x = 101/100 : 1/2

<=> x = 101/100 * 2 

<=> x = 101/50

Vậy x = 101/50 

x=100

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

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

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

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

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

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

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

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

Dấu "." là dấu nhân nha bạn

tớ chưa hiểu lắm

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}+x=100\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}+x=100\)

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

\(\left(1-\frac{1}{100}\right)+x=100\)

\(\frac{99}{100}+x=100\)

\(x=100-\frac{99}{100}=\frac{9901}{100}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}+x=100\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+....+\frac{1}{99}-\frac{1}{100}+x=100\)

\(\Rightarrow1-\frac{1}{100}+x=100\)

\(\Rightarrow\frac{99}{100}+x=100\)

\(\Rightarrow x=100-\frac{99}{100}\)

\(\Rightarrow x=\frac{1}{100}\)

~Chúc bạn hok tốt~

1/1.2+1/2.3+1/3.4+1/4.5+...+1/99.100+x=100

1/1-1/2+1/3-1/4+1/4-1/5+...+1/99-1/100+x=100

1/1-1/100+x=100

 99/100+x=100

             x=100-99/100

             x=9901/100

Chúc bạn tốt nha!

\(X-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)

\(=>X-\frac{2}{3}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

\(=>X-\frac{2}{3}=1-\frac{1}{100}\)

\(=>X-\frac{2}{3}=\frac{100}{100}-\frac{1}{100}\)

\(=>X-\frac{2}{3}=\frac{99}{100}\)

\(=>X=\frac{99}{100}+\frac{2}{3}\)

\(=>X=\frac{497}{300}\)

Lưu ý: dấu chấm thay dấu nhân

\(x-\frac{2}{3}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)

Tổng vế phải gồm : \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\)

 \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

Với vế trái, ta có : \(x-\frac{2}{3}=\frac{99}{100}\)

\(x-\frac{2}{3}=\frac{99}{100}\)

\(x=\frac{99}{100}+\frac{2}{3}\)

\(x=\frac{497}{300}\)

\(x-\frac{1}{2}-\frac{1}{6}-...-\frac{1}{9900}=200\)

\(\Leftrightarrow x-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{9900}\right)=200\)

\(\Leftrightarrow x-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=200\)

\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=200\)

\(\Leftrightarrow x-\left(1-\frac{1}{100}\right)=200\)

Ez rồi :) Tự giải tiếp

Ta có: \(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{9900}=200\)

=> \(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=200\)

=> \(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=200\)

=> \(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)=200\)

=> \(x-\left(1-\frac{1}{100}\right)=200\)

=> \(x-\frac{99}{100}=200\)

=> \(x=200+\frac{99}{100}\)

=> \(x=\frac{20099}{100}\)

Thank you