19A=192010+19/192010+1=192010+1+18/192010+1=192010+1/192010+1+18/192010+1=1+18/192010
19B=192009+19/192009+1=192009+1+18/192009+1=192009+1/192009+1+18/192009+1=1+18/192009
Vậy A<B
Xin lỗi mình chịu câu trên
Ta có A=\(\frac{19^{2009}+1}{19^{2010}+1}\) Ta có:B=\(\frac{19^{2008}+1}{19^{2009}+1}\)
19B=\(\frac{19^{2009}+19}{19^{2009}+1}\)
19A=\(\frac{19^{2010}+19}{19^{2010}+1}\) 19B=\(\frac{19^{2009}+1+18}{19^{2009}+1}\)
19A=\(\frac{19^{2010}+1+18}{19^{2010}+1}\) 19B=\(1+\frac{18}{19^{2009}+1}\)
19A=\(1+\frac{18}{19^{2010}+1}\)
Vì \(\frac{18}{19^{2010}+1}< \frac{18}{19^{2009}+1}\)nên \(19A< 19B\)
\(\Leftrightarrow A< B\)
Vậy\(A< B\)
\(\frac{19^{2008}+1}{19^{2009+1}}< \frac{19^{2008}+1+18}{19^{2009+1}+18}=\frac{19^{2008}+19}{19^{2009}+19}=\frac{19\left(19^{2007}+1\right)}{19\left(19^{2008}+1\right)}=\frac{\left(19^{2007}+1\right)}{\left(19^{2008}+1\right)}=>\frac{19^{2008}+1}{19^{2009+1}}< \frac{\left(19^{2007}+1\right)}{\left(19^{2008}+1\right)}\)
QĐMS: \(\frac{a}{b}=\frac{a.\left(b+k\right)}{b.\left(b+k\right)}=\frac{ab+ak}{b^2+bk}z\left(1\right)\)\(\frac{a+k}{b+k}=\frac{b.\left(a+k\right)}{b.\left(b+k\right)}=\frac{ba+bk}{b^2+bk}\left(2\right)\)
Bài bạn viết thiếu 1 phần là a<b. nếu a.>b thì\(\frac{a}{b}>\frac{a+k}{b+k}\)Tiếp nhé:vì a<b => ak<bk =>ab+ak<ab+bk => phân số đầu bé hơn
TH a> b thì bạn cũng giải như vậy nhé