Bài 6: So sánh phân số

Ta có :

\(A=\dfrac{2016^9+3}{2016^9-1}=\dfrac{2016^9-1+4}{2016^9-1}=\dfrac{2016^9-1}{2016^9-1}+\dfrac{4}{2016^9-1}=1+\dfrac{4}{2016^9-1}\)

\(B=\dfrac{2016^9}{2016^9-4}=\dfrac{2016^9-4+4}{2016^9-4}=\dfrac{2016^9-4}{2016^9-4}+\dfrac{4}{2016^9-4}=1+\dfrac{4}{2016^9-4}\)

Vì \(1+\dfrac{4}{2016^9-1}< 1+\dfrac{4}{2016^9-4}\Rightarrow A< B\)

Có:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{81}+\dfrac{1}{100}\)

\(A=\dfrac{1}{4}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}+\dfrac{1}{10^2}\)

Mà: \(\dfrac{1}{3^2}>\dfrac{1}{3.4}\)

\(\dfrac{1}{4^2}>\dfrac{1}{4.5}\)

...

\(\dfrac{1}{9^2}>\dfrac{1}{9.10}\)

\(\dfrac{1}{10^2}>\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(A>\dfrac{1}{4}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-0-0-...-0-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{11}\)

\(\Rightarrow A>\dfrac{65}{132}\)

Chúc bạn học tốt!

a) ta có:

11/27 <11/16=33/48<33/47

=>11/27 <33/47

a) \(\dfrac{n}{3n+1}=\dfrac{2.n}{2\left(3n+1\right)}=\dfrac{2n}{6n+2}\)

Vì \(\dfrac{2n}{6n+2}< \dfrac{2n}{6n+1}\Leftrightarrow\dfrac{n}{3n+1}< \dfrac{2n}{6n+1}\)

b) Áp dụng công thức :

\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a;b;m\in N\cdot\right)\)

Ta có :

\(B=\dfrac{10^8+1}{10^9+1}< 1\)

\(\Leftrightarrow B=\dfrac{10^8+1}{10^9+1}< \dfrac{10^8+1+9}{10^9+1+9}=\dfrac{10^8+10}{10^9+10}=\dfrac{10\left(10^7+1\right)}{10\left(10^8+1\right)}=\dfrac{10^7+1}{10^8+1}=A\)

\(\Leftrightarrow B< A\)

Ta có:

\(\dfrac{n}{3n+1}=\dfrac{2n}{2\left(3n+1\right)}=\dfrac{2n}{6n+2}\)

\(\dfrac{2n}{6n+2}< \dfrac{2n}{6n+1}\Rightarrow\dfrac{n}{3n+1}< \dfrac{2n}{6n+1}\)

Ta có:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(B=\dfrac{10^8+1}{10^9+1}< 1\)

\(\Rightarrow B< \dfrac{10^8+1+9}{10^9+1+9}\Rightarrow B< \dfrac{10^8+10}{10^9+10}\Rightarrow B< \dfrac{10\left(10^7+1\right)}{10\left(10^8+1\right)}\Rightarrow B< \dfrac{10^7+1}{10^8+1}=A\)\(\Rightarrow B< A\)

a)Ta có : \(\dfrac{1212}{1414}=\dfrac{12.101}{14.101}=\dfrac{12}{14}\)

\(\dfrac{121212}{242424}=\dfrac{12.10101}{24.10101}=\dfrac{12}{24}\)

Vậy \(\dfrac{12}{24}=\dfrac{1212}{2424}=\dfrac{121212}{242424}\)

Ta có : \(\dfrac{2424}{3535}=\dfrac{24.101}{35.101}=\dfrac{24}{35}\)

\(\dfrac{242424}{353535}=\dfrac{24.10101}{35.10101}=\dfrac{24}{35}\)

Vậy \(\dfrac{24}{35}=\dfrac{2424}{3535}=\dfrac{242424}{353535}\)

Ta có : \(\dfrac{224466}{255075}=\dfrac{22.10203}{25.10203}=\dfrac{22}{25}\)

Vậy \(\dfrac{22}{25}=\dfrac{224466}{255075}\)

Với 1 phân số bất kì \(\dfrac{a}{b}>0=>\dfrac{a+n}{b+n}>\dfrac{a}{b}\left(n>0\right)\)

\(=>\dfrac{n}{n+2}< \dfrac{n+1}{n+2+1}\)

\(=>\dfrac{n}{n+2}< \dfrac{n+1}{n+3}\)

CHÚC BẠN HỌC TỐT..........

Ta quy đồng mẫu số của hai phân số:

n+1/n+3 = (n+1).(n+2)/(n+3).(n+2) = n.(1+2)/(n+3).(n+2) =

n.3/(n+3).(n+2)

n/n+2 = n.(n+3)/(n+3).(n+2)

Ta thấy hai phân số trên đã được quy đồng mẫu nên ta sẽ so sánh hai tử: Vì n.3 < n.(n+3) nên phân số n+1/n+3 < n/n+2

. là dấu nhân, / thay cho gạch ngang của phân số nha bạn. Nếu mình làm đúng thì bạn tick nha!

Đặt \(A=\dfrac{2003.2004-1}{2003.2004}\) và \(B=\dfrac{2004.2005-1}{2004.2005}\)

Ta có : \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}\)

\(=1-\dfrac{1}{2003.2004}\)

\(B=\dfrac{2004.2005-1}{2004.2005}=\dfrac{2004.2005}{2004.2005}-\dfrac{1}{2004.2005}\)

\(=1-\dfrac{1}{2004.2005}\)

Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)

Nên \(A< B\)

Vậy \(\dfrac{2003.2004-1}{2003.2004}< \dfrac{2004.2005-1}{2004.2005}\)

~ Học tốt ~

\(\dfrac{25}{26}=\dfrac{25\cdot1010}{26\cdot1010}=\dfrac{25250}{26260}\)

Áp dụng khi \(\dfrac{a}{b}< 1\) thì \(\dfrac{a+m}{b+m}>\dfrac{a}{b}\left(m\in N^{\circledast}\right)\)

\(\dfrac{25}{26}=\dfrac{25250}{26260}< \dfrac{25250+1}{26260+1}=\dfrac{25251}{26261}\)

Vậy \(\dfrac{25}{26}< \dfrac{25251}{26261}\)

Ta có : \(\dfrac{25}{26}=\dfrac{25.101}{16.101}=\dfrac{2525}{2626}=\dfrac{25250}{26260}\)

Áp dụng công thức :

\(\dfrac{a}{b}< 1\) và \(m\in\) N, \(m\ne0\), nên

\(\dfrac{a}{b}< \dfrac{a+m}{b+m}\)

\(\Rightarrow\dfrac{25}{26}=\dfrac{25250}{26260}< \dfrac{25250+1}{26260+1}=\dfrac{25251}{26261}\)

Vậy \(\dfrac{25}{26}< \dfrac{25251}{26261}\)

~ Học tốt ~

Cho A=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\); B=\(\dfrac{2003+2004}{2004+2005}\)

Ta có: B=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)

Vì: \(\dfrac{2003}{2004+2005}< \dfrac{2003}{2004}\)

\(\dfrac{2004}{2004+2005}< \dfrac{2004}{2005}\)

=>\(\dfrac{2003}{2004+2005}+\dfrac{2004}{2004+2004}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)

=>\(\dfrac{2003+2004}{2004+2005}< \dfrac{2003}{2004}+\dfrac{2004}{2005}\)

=>B<A

Vậy B<A