Bài 6: So sánh phân số

b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)

\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)

\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)

Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)

\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)

\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)

\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)

\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)

Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)

\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)

\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)

Mà \(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)

Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)

Mình sửa kết luận

Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}< \dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=\dfrac{1}{1}-\dfrac{1}{14}=\dfrac{13}{14}\)

Ta thấy: \(\dfrac{13}{14}< 1\)

Vậy A<1

Đặt \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{13\cdot14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

\(=\dfrac{14}{14}-\dfrac{1}{14}\)

\(=\dfrac{13}{14}< 1\)

Vậy A < 1

Ta có :B= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{13.14}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{13}-\dfrac{1}{14}\)

\(=1-\dfrac{1}{14}\)

Vì \(1-\dfrac{1}{14}< 1\)

Suy ra : \(B< 1\)

A= (1/4+1/5+...+1/9)+(1/10+...+1/19)+(1/20+...+1/29)+(1/30+...+1/39)+....+(1/50+1/63)

a>1/9.6+ 1/20.10+1/30.10+... + 1/60.10

A>2/3+1/2+1/3+1/4+1/5+1/6

A>1+1/2+1/4+1/5+1/6

A>1+67/60

=> A>2+7/60

=> A>2

ahuhu ai giúp mk vs

B= \(\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)

Ta có : \(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002};\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)

\(\Rightarrow\) \(\dfrac{2000}{2001}+\dfrac{2001}{2000}>\dfrac{2000+2001}{2001+2002}\)

\(\Rightarrow\) A>B

Học tốt nha

A<B

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{217.218}\)

A = \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{217}-\dfrac{1}{218}\)

A = 1 - \(\dfrac{1}{218}\)

B = \(\dfrac{1}{110}\) + \(\dfrac{1}{111}\) + \(\dfrac{1}{112}\) + ... + \(\dfrac{1}{218}\)

Xét dãy số 110; 111; 112; ...; 218, dãy số này có số số hạng là:

         (218 - 110) : 1 + 1  =  109 (số)

Mặt khác \(\dfrac{1}{110}\) > \(\dfrac{1}{111}>\dfrac{1}{112}>...>\dfrac{1}{218}\)

⇒ B = \(\dfrac{1}{110}\) + \(\dfrac{1}{111}\) + \(\dfrac{1}{112}+...+\dfrac{1}{218}\) < \(\dfrac{1}{110}\) + \(\dfrac{1}{110}\)+ ... +\(\dfrac{1}{110}\)  

   B < \(\dfrac{1}{110}\) x 109

B  <  1 - \(\dfrac{1}{110}\)

\(\dfrac{1}{128}\) < \(\dfrac{1}{110}\) ⇒ A =  1 - \(\dfrac{1}{128}\) > 1 - \(\dfrac{1}{110}\)  > B 

A > B 

Giải:

Ta có:

\(\dfrac{10^n}{10^n+1}=\dfrac{10^n+1-1}{10^n+1}=1-\dfrac{1}{10^n+1}\)

\(\dfrac{10^n+1}{10^n+2}=\dfrac{10^n+2-1}{10^n+2}=1-\dfrac{1}{10^n+2}\)

Vì \(\dfrac{1}{10^n+1}>\dfrac{1}{10^n+2}\)

\(\Leftrightarrow-\dfrac{1}{10^n+1}< -\dfrac{1}{10^n+2}\)

\(\Leftrightarrow-\dfrac{1}{10^n+1}+1< -\dfrac{1}{10^n+2}+1\)

Hay \(1-\dfrac{1}{10^n+1}< 1-\dfrac{1}{10^n+2}\)

\(\Leftrightarrow\dfrac{10^n}{10^n+1}< \dfrac{10^n+1}{10^n+2}\)

Vậy ...

Sai đề rồi \(\dfrac{1}{13}\) chứ đâu phải \(\dfrac{1}{3}\)

Ta có: \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(S=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

\(< \dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{61}+\dfrac{1}{61}+\dfrac{1}{61}\right)\)

\(=\dfrac{1}{5}+\dfrac{1}{13}.3+\dfrac{1}{61}.3\)

\(=\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)

\(< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{3}{60}\)

\(=\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)

\(=\dfrac{10}{20}\)\(=\dfrac{1}{2}\)

Vậy S\(< \dfrac{1}{2}\) (đpcm)

a. 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

b, A =(1/101+1/102+....+1/150)+(1/151+1/152+.....+1/200)
A>1/150.50+1/200.50=1/3+1/4=7/12
b tách A thành bốn nhóm rồi cũng làm như trên,ta có
A>25/125+25/150+25/175+25/200=(1/5+1/6+1/7)+1/8
=107/210+1/8>1/2+1/8=5/8

bài này ta quy đồng mẫu rồi so sánh là xong