\(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}\)
\(2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}\)
\(=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}\)
\(=1+\dfrac{2007}{2008^{2009}+1}\)
\(B=\dfrac{2008^{2007}+1}{2008^{2008}+1}\)
\(2008\cdot B=\dfrac{2008^{2008}+2008}{2008^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2007}{2008^{2008}+1}\)
\(=1+\dfrac{2007}{2008^{2008}+1}\)
Ta có: \(2008^{2009}+1>2008^{2008}+1\)
\(\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}\)
\(\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}\)
\(\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}\)
hay \(A < B\)
#\(Toru\)
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ta có Đặt \(A=\frac{2008^{2008}+1}{2008^{2009}+1}\)
\(B=\frac{2008^{2007}+1}{2008^{2008}+1}\)
Xét A trước ta có
\(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}\)\(2008A=\frac{2008^{2009}+2008}{2008^{2009}+1}\)
\(2008A=\frac{2008^{2009}+1+2007}{2008^{2009}+1}\)suy ra \(2008A=1+\frac{2007}{2008^{2009}+1}\)
Xét B ta có
\(2008B=\frac{2008.\left(2008^{2007}+1\right)}{2008^{2008}+1}\)suy ra \(2008B=\frac{2008^{2008}+2008}{2008^{2008}+1}\)
\(2008B=\frac{2008^{2008}+1+2007}{2008^{2008}+1}\)suy ra \(2008B=1+\frac{2007}{2008^{2008}+1}\)
VÌ \(1+\frac{2007}{2008^{2009}+1}<1+\frac{2007}{2008^{2008}+1}\)
suy ra 2008A<2008B suy ra A<B
Đặt \(a=2008^{2007};\)
\(A=\frac{2008^{2008}+1}{2008^{2009}+1}=\frac{2008a+1}{2008^2.a+1};\text{ }B=\frac{2008^{2007}+1}{2008^{2008}+1}=\frac{a+1}{2008a+1}\)
Quy đồng mẫu ta có:
\(A=\frac{\left(2008a+1\right)\left(2008a+1\right)}{\left(2008^2a+1\right)\left(2008a+1\right)}=\frac{2008^2a^2+2.2008a+1}{\left(2008^2a+1\right)\left(2008a+1\right)}\)
\(B=\frac{\left(a+1\right)\left(2008^2a+1\right)}{\left(2008a+1\right)\left(2008^2a+1\right)}=\frac{2008^2a^2+\left(2008^2+1\right)a+1}{\left(2008a+1\right)\left(2008^2a+1\right)}\)
So sánh ở tử ta thấy \(2.2008<2008^2+1\)
Suy ra A < B.
ý, nếu không được dùng cách kia thì làm cách này cho chắc đi :v
Ta có: \(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}=\frac{2008^{2009}+2008}{2008^{2009}+1}=\frac{\left(2008^{2009}+1\right)+2007}{2008^{2009}+1}=1+\frac{2007}{2008^{2009}+1}\)
Lại có: \(2008B=\frac{2008\left(2008^{2007}+1\right)}{2008^{2008}+1}=\frac{2008^{2008}+2008}{2008^{2008}+1}=\frac{\left(2008^{2008}+1\right)+2007}{2008^{2008}+1}=1+\frac{2007}{2008^{2008}+1}\)
Vì 2008 < 2009 \(\Rightarrow2008^{2008}< 2008^{2009}\)\(\Rightarrow2008^{2008}+1< 2008^{2009}+1\)\(\Rightarrow\frac{2007}{2008^{2008}+1}>\frac{2007}{2008^{2009}+1}\)\(\Rightarrow1+\frac{2007}{2008^{2008}+1}>1+\frac{2007}{2008^{2009}+1}\)\(\Rightarrow2008B>2008A\)\(\Rightarrow B>A\)
Vì A <1 , B < 1
Nên ta có: \(A=\frac{2008^{2008}+1}{2008^{2009}+1}< \frac{2008^{2008}+1+2007}{2008^{2009}+1+2007}=\frac{2008^{2008}+2008}{2008^{2009}+2008}=\frac{2008\left(2008^{2007}+1\right)}{2008\left(2008^{2008}+1\right)}=\frac{2008^{2007}+1}{2008^{2008}+1}=B\)
Nhận xét: nếu 0<�<�0<x<y và �>0a>0 thì 0<��<��0<ax<ay và ��+��<��+��⇒�(�+�)<�(�+�)⇒��<�+��+�xy+ax<xy+ay⇒x(y+a)<y(x+a)⇒yx<y+ax+a.
Do đó, (20082008+1)(20082009+1)<(20082008+1)+2007(20082009+1)+2007⇒20082008+120082009+1<2008(20082007+1)2008(20082008+1)(20082009+1)(20082008+1)<(20082009+1)+2007(20082008+1)+2007⇒20082009+120082008+1<2008(20082008+1)2008(
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Nhận xét: nếu 0<�<�0<x<y và �>0a>0 thì 0<��<��0<ax<ay và ��+��<��+��⇒�(�+�)<�(�+�)⇒��<�+��+�xy+ax<xy+ay⇒x(y+a)<y(x+a)⇒yx<y+ax+a.
Do đó, (20082008+1)(20082009+1)<(20082008+1)+2007(20082009+1)+2007⇒20082008+120082009+1<2008(20082007+1)2008(20082008+1)(20082009+1)(20082008+1)<(20082009+1)+2007(20082008+1)+2007⇒20082009+120082008+1<2008(20082008+1)2008(
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Câu hỏi trên có đáp án là A<B