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1/101>1/150 
1/102>1/150 
1/103>1/150 
.... 
1/150=1/150 
Tất cả có 50 dữ kiện 
Vậy 1/101+1/102+...+1/150>50/150=1/3 (1) 

Tiếp theo 
1/151>1/200 
1/152>1/200 
... 
1/200=1/200 
Tương tự trên, thì :
1/151+......+1/200>50/200=1/4 (2) 

Cộng (1) và (2), thì A>(1/3+1/4)=7/12 \(\left(ĐPCM\right)\).

Ta có :

\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}=VP\left(đpcm\right)\)

Xét :

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)

Thêm \(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\)vào mỗi vế ta có

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

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

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

\(\RightarrowĐPCM\)

Tách A thành 2 nhóm A₁ , A₂

A₁ = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)

A₂ = \(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\)

\(\Rightarrow\)A = A₁ + A₂ > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

\(A>\left(\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\right)+\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)\) (mỗi ngoặc có 50 số hạng)

\(;A>\left(\frac{1}{150}.50\right)+\left(\frac{1}{200}.50\right)=50.\left(\frac{1}{150}+\frac{1}{200}\right)=50.\frac{7}{600}=\frac{7}{12}\)

Ta có : \(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{200}=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

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

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

\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)\(\left(đpcm\right)\)

 ta có 
1/101 > 1/150
1/102> 1/150
...>1/150
1/150 = 1/150
=> 1/101 + 1/102 + .... + 1/150 > 1/150 +1/150+....+1/150(50 số hạng )= 1/3
ta có
1/151 >1/200
1/152 > 1/200
..>1/200
1/200 = 1/200
=> 1/151 + 1/152+....+1/200 > 1/200+1/200+ ...+1/200( 50 số hạng) = 1/4
==> 1/101 + 1/102+....+1/200 > 1/3 +1/4
==> A > 7/12

A>7/12

Biến đổi vế phải của đẳng thức :

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

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{100}\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right]\)

\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{100}\)

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)+\left(\frac{1}{101}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{199}+\frac{1}{200}\) (ĐPCM)

Ta có : 1 - 1/2 + 1/3 - 1/4 + ....- 1/200

= (1 + 1/3 + 1/5 + ....+ 1/199) - ( 1/2 + 1/4 + 1/6 + .... + 1/200)

= ( 1 + 1/3 +...+ 1/199) + (1/2 +1/4 + ...+ 1/200) - 2(1/2+1/4+...+ 1/200)

= (1+1/2+1/3+....+1/199 + 1/200) - (1 +1/2 +1/3 +....+1/100)

= 1/101 + 1/102+ 1/103 + .... + 1/200

chúc bạn học tốt!!!!!!!