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Mik nghĩ bài này nếu quy đồng mẫu của cả A và B thì chúng sẽ có cùng chung mẫu vì đều là 10^2005 và 10^2006.
Như vậy nếu cộng tử số của cả A và B thì A và B sẽ bằng nhau.=>A và B bằng nhau
Đây chỉ là suy nghĩ của mình thôi vì mik chx lm bài này bao h
tham khảo
https://hoc24.vn/cau-hoi/so-sanh-ko-qua-quy-donga-7102005-15102006b-15102005-7102006.78462087582#:~:text=%3D%3EA%3D,%3D%3EA%3CB
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:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}< 1\)
\(A< \dfrac{2006^{2006}+1+2005}{2006^{2007}+1+2005}\Rightarrow A< \dfrac{2006^{2006}+2006}{2006^{2007}+2006}\Rightarrow A< \dfrac{2006\left(2006^{2005}+1\right)}{2006\left(2006^{2006}+1\right)}\Rightarrow A< \dfrac{2006^{2005}+1}{2006^{2006}+1}=B\)\(A< B\)
\(C=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(1+\dfrac{2005}{2}\right)+\left(1+\dfrac{2004}{3}\right)+...+\left(1+\dfrac{1}{2006}\right)+1}\)
\(=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2007}}=\dfrac{2006}{2007}\)
\(A=\frac{1000^{2004}+1}{1000^{2005}+1}\)
=> \(1000A=\frac{1000^{2005}+1000}{1000^{2005}+1}=1+\frac{999}{1000^{2005}+1}\)
\(B=\frac{1000^{2005}+1}{1000^{2006}+1}\)
=> \(1000A=\frac{1000^{2006}+1000}{1000^{2006}+1}=1+\frac{999}{1000^{2006}+1}\)
Vì: \(1000^{2006}+1>1000^{2005}+1\)
=> \(\frac{999}{1000^{2006}+1}< \frac{99}{1000^{2005}+1}\)
=> \(1+\frac{999}{1000^{2006}+1}< 1+\frac{99}{1000^{2005}+1}\)
=> 1000B < 1000A
=> B < A
Ta có :
\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2006}}\)
\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2005}}\)
Lại có :
\(-\dfrac{8}{10^{2006}}>\dfrac{-8}{10^{2005}}\Leftrightarrow M>N\)
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) 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