Question

BT7: So sánh

1) \(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}\)và \(B=\dfrac{2005^{2004}+1}{2005^{2005}+1}\)

Answer 1

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}< 1\)

\(A< \dfrac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\Rightarrow A< \dfrac{2005^{2005}+2005}{2005^{2006}+2005}\Rightarrow A< \dfrac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\Rightarrow A< \dfrac{2005^{2004}+1}{2005^{2005}+1}=B\)

\(A< B\)

Answer 2

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 \(A< B\)