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Ta có : \(A=\frac{2009.2009+2008}{2009.2009+2009}\)
\(=1-\frac{1}{2009.2009+2009}\)
\(B=\frac{2009.2009+2009}{2009.2009+2010}\)
\(=1-\frac{1}{2009.2009.2010}\)
Mà \(-\frac{1}{2009.2009+2009}< -\frac{1}{2009.2009.2010}\)
=> \(\frac{2009.2009+2008}{2009.2009+2009}< \frac{2009.2009+2009}{2009.2009.2010}\) => A < B
b) Áp dụng tính chất
\(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Ta có: \(B=\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=\frac{10^{16}+10}{10^{17}+10}=\frac{10.\left(10^{15}+1\right)}{10.\left(10^{16}+1\right)}=\frac{10^{15}+1}{10^{16}+1}=A\)
\(\Rightarrow B< A\)
\(B< 1\Rightarrow\frac{10^{16}+1}{10^{17}+1}< \frac{10^{16}+1+9}{10^{17}+1+9}=\frac{10^{16}+10}{10^{17}+10}=\frac{10\left(10^{15}+1\right)}{10\left(10^{16}+1\right)}=\frac{10^{15}+1}{10^{16}+1}=A\)
\(\Rightarrow A>B\)
\(A=\frac{17^{18}+1}{17^{19}+1}\)
\(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)
\(B=\frac{17^{17}+1}{17^{18}+1}\)
\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{\left(17^{18}+1\right)+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)
\(\text{Vì}\)\(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)
\(\Leftrightarrow17A< 17B\)
\(\Leftrightarrow A< B\)
Trả lời
\(17A=\frac{\left(17^{18}+1\right)17}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)
\(17B=\frac{\left(17^{17}+1\right)17}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)
Vì \(17^{19}+1>17^{18}+1\)
\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)
\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)
\(\Rightarrow B>A\)
cu lay phep tinh nay tru phep tinh kia hk ra thi nt hoi mink
\(b)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Vậy \(\frac{2009^{2009}+1}{2009^{2010}+1}>\frac{2009^{1010}-2}{2009^{2011}-2}\)
Chúc bạn học tốt ~
Àk mình còn thiếu một điều kiện nữa xin lỗi nhé :
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Bạn thêm vào nhé
17A = \(\frac{17^{2009}+17}{17^{2009}+1}=1+\frac{16}{17^{2009}+1}\)
17B = \(\frac{17^{2010}+17}{17^{2010}+1}=1+\frac{16}{17^{2010}+1}\)
mà \(\frac{16}{17^{2009}+1}>\frac{16}{17^{2010}+1}\)
=> A > B
B < 17 ^ 2009 + 1 + 16 / 17^2010 + 1+16 = 17^2009 + 17 / 17^2010 + 17 = 17(17^2008 + 1) / 17(17^2009+1) = 17^2008 + 1 / 17^2009 + 1 =A
=> B < A
****** k mk nha!