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Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{2005^{2014}+1}{2005^{2015}+1}< 1\\B=\dfrac{2005^{2015}+1}{2005^{2016}+1}< 1\end{matrix}\right.\)
\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2015}+1+2004}{2005^{2016}+1+2004}\)
\(=\dfrac{2005^{2015}+2005}{2005^{2016}+2005}=\dfrac{2005\left(2005^{2014}+1\right)}{2005\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
\(\Rightarrow\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(B< A\)
Hay \(A>B\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}< 1\)
\(A< \dfrac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\Rightarrow A< \dfrac{2005^{2005}+2005}{2005^{2006}+2005}\Rightarrow A< \dfrac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\Rightarrow A< \dfrac{2005^{2004}+1}{2005^{2005}+1}=B\)
\(A< B\)
Ta có : A = \(\dfrac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\)A = \(\dfrac{\left(2005^{2005}+1\right).2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005A=\dfrac{2005^{2006}+1}{2005^{2006}+1}+\dfrac{2004}{2005^{2006}+1}\)
\(2005A=1+\dfrac{2004}{2005^{2006}+1}\)
Tương tự như vậy với \(B\) ta đc
\(2005B=1+\dfrac{2004}{2005^{2005}+1}\)
Vì \(2005^{2006}+1>2005^{2005}+1\)
\(=>\) \(1+\dfrac{2004}{2005^{2006}+1}\)\(< \)\(1+\dfrac{2004}{2005^{2005}+1}\)
\(=>\)\(2005A< 2005B\)
\(=>\)\(A< B\)
Vậy \(A< B\)
\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)
\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)
A=2016/2017+2017/2018
Do 2016/2017<1,2017/2018<1=> A<2 Hay A<B
\(\frac{100^{2015}+1}{100^{2015}+1}=1\)
\(\frac{100^{2016}+1}{100^{2016}+1}=1\)
Vì 1 = 1 nên \(\frac{100^{2015}+1}{100^{2015}+1}=\frac{100^{2016}+1}{100^{2016}+1}\)
à mình nhìn nhầm đề
Mình giải nha
Đặt \(A=\frac{100^{2015}+1}{100^{2005}+1}\Rightarrow\frac{A}{100^{10}}=\frac{100^{2015}+1}{100^{2015}+100^{10}}=\frac{100^{2015}+100^{10}-999}{100^{2015}+100^{10}}=1-\frac{999}{100^{2015}+100^{10}}\)
Đặt \(B=\frac{100^{2016}+1}{100^{2006}+1}\Rightarrow\frac{B}{100^{10}}=\frac{100^{2016}+100^{10}-999}{100^{2016}+100^{10}}=1-\frac{999}{100^{2016}+100^{10}}\)
\(1-\frac{999}{100^{2015}+100^{10}}< 1-\frac{999}{100^{2016}+100^{10}}\Rightarrow A< B\)
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