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

Ta có:

\(\dfrac{1}{101}>\dfrac{1}{200}\)

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

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

...

\(\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)

\(=\dfrac{1}{200}.100\)

\(=\dfrac{1}{2}\)

Mà \(\dfrac{1}{2}< 1\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}< 1\).

Ta có :  1/101 > 1/150 , 1/102 > 1/150 , ... , 1/149 > 1/150 

=> 1/101 + 1/102 + ...+ 1/149 + 1/150 > 1/150 x 50 ( có tất cả 50 số ) 

=> A > 50/150

=> A > 1/3

Vậy A > 1/3

Chúc bạn học giỏi !!! TK mình nha !!! 

Bạn tham khảo nhé 

Ta có : 

\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}=50.\frac{1}{150}=\frac{50}{150}=\frac{1}{3}\)

Vậy \(A>\frac{1}{3}\)

Chúc bạn học tốt ~

sakura ???

De sai o dau phai hok ban. Phien ban xem lai giup.Toi mik giai cho

-3-2x+17=12-3x

-2x+3x=12-17+3

x=-2

b.43-2x-2=-48+x

  -2x-x = -48+2-43

   -3x=-89

    x=-89/-3

c  -253-2x=102+19-x

        -2x+x=102+19+253

         -x=374

             x=-374

\(A=\left(2^{101}-2^{100}+2^{99}-2^{98}+....+2^2-2+1\right):\left(2^{102}-1\right)\)

\(S=2^{100}+2^{98}+...+2^2+2^0\text{ do đó: }4S=2^{102}+2^{100}+...+2^4+2^2\Rightarrow4S-S=2^{102}-1\Rightarrow A=S:\left(2^{102}-1\right)=\dfrac{1}{3}\)