A=1/5.7 + 1/7.9 + ..... + 1/99.101

A=1/5 - 1/7 + 1/7 - 1/9 + ..... + 1/99 - 1/101

A=1/5 - 1/101 = 1/116

=> A không phải là số tự nhiên

\(A=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+...+\frac{19}{9^2\cdot10^2}\\ A=\frac{3}{1\cdot4}+\frac{5}{4\cdot9}+...+\frac{19}{81\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)

Ta thấy \(0< \frac{99}{100}< 1\)

\(\Rightarrow0< A< 1\)

\(\Rightarrow A\notin N\)

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(\Rightarrow A=\frac{2^2-1}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)

\(\Rightarrow A=\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)

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

\(\Rightarrow A=\frac{1}{1^2}-\frac{1}{10^2}\)

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

\(\Rightarrow A=\frac{99}{100}.\)

Vì \(0< \frac{99}{100}< 1.\)

\(\Rightarrow0< A< 1.\)

\(\Rightarrow A\notin N\left(đpcm\right).\)

Chúc bạn học tốt!

\(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+.....+\frac{1}{101}\)

\(=\frac{1}{2+3}+\frac{1}{3+4}+\frac{1}{4+5}+....+\frac{1}{50+51}\)

Anh quên mất đoạn sau rồi , nhưng hình như đến đây kl là được rồi đấy

\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)

\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)

Vì  \(\frac{2}{6}>\frac{2}{12};\frac{2}{8}>\frac{2}{12};\frac{2}{10}>\frac{2}{12};...;\frac{1}{11}>\frac{2}{12}\)

\(\Rightarrow E=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>6.\frac{2}{12}=1\) \(\left(1\right)\)

Vì \(\frac{2}{8}< \frac{2}{6};\frac{2}{10}< \frac{2}{6};...;\frac{2}{11}< \frac{2}{6}\)

\(\Rightarrow E=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< 6.\frac{2}{6}=2\) \(\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow1< E< 2\Rightarrow E\notin Z\)(đpcm)

\(B=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{100}}\)

\(\Rightarrow B< \frac{2}{2\sqrt{1}}+\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{99}+\sqrt{100}}\)

\(\Rightarrow B< 1+2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)

\(\Rightarrow B< 1+2\left(\sqrt{100}-\sqrt{1}\right)\Rightarrow B< 19\)

Tương tự:

\(B>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{101}-\sqrt{100}}\)

\(\Rightarrow B>2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{101}-\sqrt{100}\right)\)

\(\Rightarrow B>2\left(\sqrt{101}-\sqrt{1}\right)>2\left(\sqrt{100}-\sqrt{1}\right)=18\)

\(\Rightarrow18< B< 19\Rightarrow B\) không phải là số tự nhiên

bạn giải ra hộ mình nhé !

a) M>N

b)M*N=1/101

c)bỏ cuộc 

BẠN giải hộ mình nhé !

Ta có : 

\(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)

\(A=\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}\right)\)

\(A>\left(\frac{1}{150}+\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\right)+\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)\)

\(A>50.\frac{1}{150}+50\frac{1}{200}\)

\(A>\frac{50}{150}+\frac{50}{200}\)

\(A>\frac{1}{3}+\frac{1}{4}\)

\(A>\frac{7}{12}\)

Vậy \(A>\frac{7}{12}\)

Chúc bạn học tốt ~ 

Ta có:\(\frac{1}{101}>\frac{1}{200}\)

          \(\frac{1}{102}>\frac{1}{200}\)

           \(\frac{1}{103}>\frac{1}{200}\)

A=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100\)

hay A>\(\frac{7}{12}\)

A=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{200}.100\)

hay A>\(\frac{5}{8}\)

mình ko biết có đúng ko bạn xem kĩ nhé

bn làm hơi tắt