Ta có;A= 1/101^2+1/102^2+1/103^2+1/104^2+1/105^2 
A>1/(100x101)+1/(101x102)+1/(102x103)+... Vì cùng tử mẫu nhỏ hơn thì lớn hơ 
A>1/100-1/101+1/101-1/102+1/102-1/103+... 
A>1/100-1/105=1/2100=1/(2^2.3.5^2.7)=B 
Vậy A>B

Ta có:A= 1/101^2+1/102^2+1/103^2+1/104^2+1/105^2 
         A>1/(100x101)+1/(101x102)+1/(102x103)+... 

Vì cùng tử mẫu nhỏ hơn thì lớn hơ 
         A>1/100-1/101+1/101-1/102+1/102-1/103+... 
        A>1/100-1/105=1/2100=1/(2^2.3.5^2.7)=B 
=>Vậy A>B

Không biết

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

x thuoc R

\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)

\(3A=3^{102}+3^{103}+3^{104}+...+3^{201}\)

\(3A-A=\left(3^{102}+3^{103}+3^{104}+3^{201}\right)-\left(3^{101}+3^{102}+3^{103}+...+3^{201}\right)\)

\(2A=3^{201}-3^{101}\)

\(2A=3^{100}\)

\(\Rightarrow A=3^{100}:2\)

\(A=3^{101}+3^{102}+3^{103}+...+3^{200}\)

\(A=3^{101}+3^{102}+3^{103}+3^{104}+...+3^{197}+3^{198}+3^{199}+3^{200}\)

\(A=3^{100}\left(3+3^2+3^3+3^4\right)+...+3^{196}\left(3+3^2+3^3+3^4\right)\)

\(A=120\left(3^{100}+...+3^{196}\right)⋮120\)

Ta thấy tổng trên có 50 số hạng .

Ta có:

1/101>1/150

1/102>1/150

...

1/149>1/150

1/150=1/150

=>1/101+1/102+...+1/149+1/150>1/150+1/150+...+1/150

                                                 ---50 số hạng 1/150-------

=>1/101+1/102+...+1/149+1/150>1/150.50

=>1/101+1/102+...+1/149+1/150>50/150

=>1/101+1/102+...+1/149+1/150>1/3

an vo cai nay la vo tra loi

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C = 1 101 + 1 102 + ... + 1 200 C = ( 1 101 + 1 102 + ... + 1 120 ) + ( 1 121 + 1 122 + . . + 1 150 ) + ( 1 151 + 1 152 + … + 1 180 ) + ( 1 181 + 1 182 + ... + 1 200 ) C > 20 . 1 120 + 30 . 1 150 + 30 . 1 180 + 20 . 1 200 C > 1 6 + 1 5 + 1 6 + 1 10 C > ( 1 6 + 1 6 ) + ( 1 5 + 1 10 ) C > 1 3 + 3 10 C > 19 30 = 76 120 C > 75 120 + 1 120 C > 5 8 + 1 120 Vậy C > 5 8 còn lại giống trên