Bài 6: So sánh phân số

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 ...

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 

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

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

\(A)-\dfrac{7}{18}\) và \(-\dfrac{5}{12}\)

\(+)-\dfrac{7}{18}=-\dfrac{14}{36};-\dfrac{5}{12}=-\dfrac{15}{36}\)

\(+)14< 15\Rightarrow-\dfrac{14}{36}>-\dfrac{15}{36}\Rightarrow-\dfrac{7}{18}>-\dfrac{5}{15}\)

Vậy \(-\dfrac{7}{18}>-\dfrac{5}{12}\)

\(B)\dfrac{123}{-124}\) và \(\dfrac{-310}{311}\)

Ta có: \(1-\dfrac{123}{124}=\dfrac{1}{124}\)\(;1-\dfrac{310}{311}=\dfrac{1}{311}\)

Có: \(\dfrac{1}{124}>\dfrac{1}{311}\Leftrightarrow1-\dfrac{123}{124}>1-\dfrac{310}{311}\)\(\Leftrightarrow1+\left(-\dfrac{123}{124}\right)>1+\left(-\dfrac{310}{311}\right)\Leftrightarrow-\dfrac{123}{124}>-\dfrac{310}{311}\)

Hay \(\dfrac{123}{-124}>\dfrac{-310}{311}\)

Vậy \(\dfrac{123}{-124}>\dfrac{-310}{311}\)

207/245 nhỏ hơn 2525/2829

Sắp xếp: \(-\frac{7}{9};\frac{3}{2};-\frac{7}{5};0;\frac{-4}{-5};\frac{9}{11}\)

Quy đồng:

\(-\frac{7}{9}=-\frac{770}{990};\frac{3}{2}=\frac{1485}{990};-\frac{7}{5}=-\frac{1386}{990}\) ; \(0=\frac{0}{990};\frac{4}{5}=\frac{792}{990};\frac{9}{11}=\frac{810}{990}\)

=> \(-\frac{1386}{990};-\frac{770}{990};0;\frac{792}{990};\frac{810}{990};\frac{1485}{990}\)

Sắp xếp theo thứ tự tăng dần: \(-\frac{7}{5};-\frac{7}{9};0;\frac{4}{5};\frac{9}{11};\frac{3}{2}\)

Mẫu số chung to quá @@

a: 255/256=1-1/256

1123/1124=1-1/1124

mà 1/256>1/1124

nên -255/256>-1123/1124

b: 5/14=100/280

-3/40=-21/280

-13/140=-26/280

=>-13/140<-3/40<5/14