phân số là gì vậy bạn

\(A=\frac{10^8+1}{10^9+1}=\frac{1}{10}\left(\frac{10^9+10}{10^9+1}\right)=\frac{1}{10}\left(1+\frac{9}{10^9+1}\right)\)

\(B=\frac{10^9+1}{10^{10}+1}=\frac{1}{10}\left(\frac{10^{10}+10}{10^{10}+1}\right)=\frac{1}{10}\left(1+\frac{9}{10^{10}+1}\right)\)

\(\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\)

\(\Rightarrow A>B\)

Đặt \(M=\frac{10^8+1}{10^9+1}\) và \(N=\frac{10^9+1}{10^{10}+1}\)

Có : \(M=\frac{10^8+1}{10^9+1}\)

\(\Rightarrow10M=\frac{10^9+10}{10^9+1}=\frac{10^9+1+9}{10^9+1}=1+\frac{9}{10^9+1}\)

Lại có : \(N=\frac{10^9+1}{10^{10}+1}\)

\(\Rightarrow10N=\frac{10^{10}+10}{10^{10}+1}=\frac{10^{10}+1+9}{10^{10}+1}=1+\frac{9}{10^{10}+1}\)

Vì \(\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\) nên \(1+\frac{9}{10^9+1}>1+\frac{9}{10^{10}+1}\)

\(\Rightarrow10M>10N\Rightarrow M>N\)

Vậy M > N.

yêu cầu so sánh 2 phân số

Đặt \(A=\frac{10^{2006}+9}{10^{2007}+9}\)

\(\Rightarrow10A=\frac{10^{2007}+90}{10^{2007}+9}=1+\frac{81}{10^{2007}+9}\)

\(\frac{10^{2007}+9}{10^{2008}+9}=B\)

\(\Rightarrow10B=\frac{10^{2008}+90}{10^{2008}+9}=1+\frac{81}{10^{2008}+9}\)

Vì\(10A>10B\Rightarrow A>B\)

Đặt \(A=\frac{10^{2006}+9}{10^{2007}+9}\) và \(B=\frac{10^{2007}+9}{10^{2008}+9}\)

* Cách 1 : 

Ta có : 

\(10A=\frac{10^{2007}+90}{10^{2007}+9}=\frac{10^{2007}+9+81}{10^{2007}+9}=\frac{10^{2007}+9}{10^{2007}+9}+\frac{81}{10^{2007}+9}=1+\frac{81}{10^{2007}+9}\)

\(10B=\frac{10^{2008}+90}{10^{2008}+9}=\frac{10^{2008}+9+81}{10^{2008}+9}=\frac{10^{2008}+9}{10^{2008}+9}+\frac{81}{10^{2008}+9}=1+\frac{81}{10^{2008}+9}\)

Vì \(\frac{81}{10^{2007}+9}>\frac{81}{10^{2008}+9}\) nên \(1+\frac{81}{10^{2007}+9}>1+\frac{81}{10^{2008}+9}\)

Vậy \(A>B\)

* Cách 2 : 

Ta có công thức : 

\(\frac{a}{b}< \frac{a+m}{b+m}\) \(\left(\frac{a}{b}< 1;a,b,m\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(B=\frac{10^{2007}+9}{10^{2008}+9}< \frac{10^{2007}+9+1}{10^{2008}+9+1}=\frac{10^{2007}+10}{10^{2008}+10}=\frac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\frac{10^{2006}+1}{10^{2007}+1}=A\)

Ta thấy \(B< A\) hay \(A>B\)

Vậy \(A>B\)

Chúc bạn học tốt ~ 

ta có -9\10^2011=-9\10^2011

mà -19\10^2011>-19\10^2011

nên A>B

****

a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A

                                                       Vậy A > B

b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A

                                                                         Vậy A < B.

                    NHỚ K CHO MK VỚI NHÉ !!!!!!!!

a A lon hon B

a) 

\(9A=\frac{9^{19}+9}{9^{19}+1}=\frac{9^{19}+1+8}{9^{19}+1}=1+\frac{8}{9^{19}+1}\)

\(9A=\frac{9^{20}+9}{9^{20}+1}=\frac{9^{20}+1+8}{9^{20}+1}=1+\frac{8}{9^{20}+1}\)

ta thấy \(9^{19}+1< 9^{20}+1\Rightarrow\frac{8}{9^{19}+1}>\frac{8}{9^{20}+1}\)

\(\Rightarrow9A>9B\Rightarrow A>B\)

A=\(\frac{-199}{10^{2011}}\)

B=\(\frac{-109}{10^{2011}}\)

Dễ dàng so sánh được A<B

A=-9/10²⁰¹¹+(-19/10²⁰¹⁰)

B=-9/10²⁰¹⁰+(-19/10²⁰¹¹)

Vì -9/10²⁰¹¹>(-19/10²⁰¹¹) và -9/10²⁰¹¹-(-19/10²⁰¹¹)=10/10²⁰¹¹

-19/10²⁰¹⁰<(-9/10²⁰¹⁰) và -9/10²⁰¹⁰-(-19/10²⁰¹⁰)=10/10²⁰¹⁰

mà 10/10²⁰¹¹<10/10²⁰¹⁰ nên suy ra B>A

Áp dụng a/b < 1 => a/b < a+m/b+m (a;b;m thuộc N*)

Ta có:

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

\(B< \frac{10^9+10}{10^{10}+10}\)

\(B< \frac{10.\left(10^8+1\right)}{10.\left(10^9+1\right)}\)

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

=> B < A

Ta có:

\(10A=\frac{10\left(10^8+1\right)}{10^9+1}=\frac{10^9+10}{10^9+1}=\frac{10^9+1+9}{10^9+1}=\frac{10^9+1}{10^9+1}+\frac{9}{10^9+1}=1+\frac{9}{10^9+1}\)

tương tự với B ta có:\(10B=1+\frac{9}{10^{10}+1}\)

Vì 10⁹+1<10¹⁰+1 \(\Rightarrow\frac{9}{10^9+1}>\frac{9}{10^{10}+1}\)

\(\Rightarrow1+\frac{9}{10^9+1}>1+\frac{9}{10^{10}+1}\)

\(\Rightarrow10A>10B\Leftrightarrow A>B\)