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Ta có:
\(2006A=\dfrac{2006^{2007}+2016}{2006^{2007}+1}=1+\dfrac{2005}{2006^{2007}+1}\)
\(2006B=\dfrac{2006^{2006}+2006}{2006^{2006}+1}=1+\dfrac{2005}{2006^{2006}+1}\)
Do \(\dfrac{2005}{2006^{2006}+1}>\dfrac{2005}{2006^{2007}+1}\Rightarrow1+\dfrac{2005}{2006^{2006}+1}>1+\dfrac{2005}{2006^{2007}+1}\)
\(\Rightarrow2006A< 2006B\Rightarrow A< B\)
Mình sẽ giải cách ngắn hơn cách bạn đạt nha:
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}< 1\)
\(A< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}\Rightarrow A< \dfrac{2006^{2006}+2006}{2006^{2007}+2006}\Rightarrow A< \dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}\Rightarrow A< \dfrac{2006^{2005}+1}{2006^{2006}+1}=B\)\(A< B\)
a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
Áp dụng Bất đẳng thức :
\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}=\dfrac{2006^{2006}+2006}{2006^{2007}+2006}=\dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
\(\Leftrightarrow\dfrac{2006^{2006}+1}{2006^{2007}+1}< \dfrac{2006^{2005}+1}{2006^{2006}+1}\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
Ta có: \(C=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{1+\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)}\)
\(=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}}\)
\(=\dfrac{2006}{2007}\)
A=2006^2005+1/2006^2006+1
B=2006^2006+1/2006^2007+1
Có : 2006A = 2006^2006+2006/2006^2006+1
= 1 + 2005/2006^2006+1 2006B
= 2006^2007+2006/2006^2007+1
= 1 + 2005/2006^2007+1
Vì : 2006^2006 < 2006^2007
=> 2006^2006+1 < 2006^2007+1
=> 2005/2006^2006+1 > 2005/2006^2007+1
=> 2016A > 2016B
=> A>B
Ta có:
\(A=\frac{2006^{2005}+1}{2006^{2006}+1}\)
\(\Rightarrow2006A=\frac{2006^{2006}+2006}{2006^{2006}+1}=\frac{\left(2006^{2006}+1\right)+2005}{2006^{2006}+1}=1+\frac{2005}{2006^{2006}+1}\)
Ta lại có:
\(B=\frac{2006^{2006}+1}{2006^{2007}+1}\)
\(\Rightarrow2006B=\frac{2006^{2007}+2006}{2006^{2007}+1}=\frac{\left(2006^{2007}+1\right)+2005}{2006^{2007}+1}=1+\frac{2005}{2006^{2007+1}}\)
Ta thấy:
\(\frac{2005}{2006^{2006}+1}>\frac{2005}{2006^{2007}+1}\Rightarrow2006A>2006B\Rightarrow A>B\)
Vậy A>B.
Ai k mình, mình k lại.
- Mình dùng cách lớp 8 để làm câu b được không :)?
ko :)