A= 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 

=1/(1.2)+1/(2.3)+1/(3.4)+1/(4.5) +1/(5.6)+1/(6.7)+1/(7.8) +1/(8.9)+1/(9.10) 

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5.+1/5-1/6... +1/9-1/10 

=1-1/10 

=9/10

Me too ! 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}=x-\left(2+4+..+100\right)\)

Gọi \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)là \(A\)

         \(\left(2+4+...+100\right)\)là \(B\). Ta có :  

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\)

\(A=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

Số số hạng của \(B\)là: \(\left(100-2\right)\div2+1=50\)

Tổng của \(B\)là: \(\left(2+100\right)\times50\div2=2550\)

\(\Rightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=x-\left(2+4+..+100\right)=\frac{9}{10}=x-2550\)

\(\Rightarrow x=2550+\frac{9}{10}=2550+0,9=2550,9\)

1/100 - 1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 - 1/6 - 1/2 = -0,833

giúp

=> (1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9 + 1/9.10) : x = 9/20

=> (1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9 + 1/9 - 1/10) : x = 9/20

=> (1 - 1/10) : x = 9/20

=> 9/10 : x = 9/20

X = 9/10 : 9/20 = 2

hhhhhhijijkn

\(2\times x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)

\(2\times x-2\times\frac{1}{12}+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)

\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)

\(2\times\left(x-\frac{1}{12}\right)+\left(\frac{1}{3}-\frac{1}{10}\right)=\frac{3}{11}\)

\(2\times\left(x-\frac{1}{12}\right)+\frac{7}{30}=\frac{3}{11}\)

\(2\times\left(x-\frac{1}{12}\right)=\frac{13}{330}\)

\(x-\frac{1}{12}=\frac{13}{660}\)

\(x=\frac{17}{165}\)

\(2x-\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=\frac{3}{11}\)

\(\Rightarrow2x-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{3}{11}\)

\(\Rightarrow2x-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{3}{11}\)

\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{3}{11}\)

\(\Rightarrow2x-\left(\frac{1}{2}-\frac{1}{10}\right)=\frac{3}{11}\)

\(\Rightarrow2x-\frac{2}{5}=\frac{3}{11}\)

\(\Rightarrow2x=\frac{3}{11}+\frac{2}{5}\)

\(\Rightarrow2x=\frac{37}{55}\)

\(\Rightarrow x=\frac{37}{55}:2\)

\(\Rightarrow x=\frac{37}{110}\)

Vậy  \(x=\frac{37}{110}\)

_Chúc bạn học tốt_

Bài 1:

a) A=1+2²+2⁴+.................+2¹⁰⁰

2A=(1+2²+2⁴+.................+2¹⁰⁰)

2A=2+2³+...+2¹⁰¹

2A-A=(2+2³+...+2¹⁰¹)-(1+2²+2⁴+.................+2¹⁰⁰)

A=2¹⁰¹-1

b)bạn tự làm

c) C=-1/90-1/72-1/50-1/42-1/30-1/20-1/12-1/6-1/2

\(=-\left(\frac{1}{90}+\frac{1}{72}+...+\frac{1}{2}\right)\)

\(=-\left(\frac{1}{10.9}+\frac{1}{9.8}+...+\frac{1}{2.1}\right)\)

\(=-\left(\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+...+\frac{1}{2}-1\right)\)

\(=-\left(\frac{1}{10}-1\right)\)

\(=-\left(-\frac{9}{10}\right)=\frac{9}{10}\)

Bài 2:

cứ tính lần lượt là ra

làm được câu a thui

\(\frac{1}{100}=\frac{1}{10.10}\);\(\frac{1}{90}=\frac{1}{9.10}\);...

Suy ra \(\frac{1}{10.10}-\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-...-\frac{1}{1.2}\)

\(\frac{1}{10}-\frac{1}{10}-\frac{1}{10}-\frac{1}{9}-...-1-\frac{1}{2}\)

\(\frac{1}{10}-\frac{1}{2}\)

\(-\frac{4}{10}\)

Thanks bạn nha

sai rồi đừng chép

Mình nghĩ bài 1 đề sai .

bài 1 và bài 2 sai đề