Toán so sánh:A=2006^2006+1/2007^2007+1 và B=2006^2005+1/2006^2006+1 13/07/2021 By Elliana so sánh:A=2006^2006+1/2007^2007+1 và B=2006^2005+1/2006^2006+1
Ta có : A = 2016 ^ 2016 + 1 / 2006 ^ 2007 + 1 < 2006 ^ 2006 + 2005 + 1 / 2006 ^ 2007 + 2005 + 1 = 2006 ^ 2006 + 2006 / 2006 ^ 2007 + 2006 = 2006 . ( 2005 + 1 ) / 2006 . ( 2006 + 1) = 2006 ^ 2005 +1 / 2006 ^ 2006 + 1 = B => A < B Chúc bn học giỏi nha ! ( sorry , mk viết hơi khó hiểu) Trả lời
Đáp án: \[A < B\] Giải thích các bước giải: Ta có: \(\begin{array}{l}A = \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}}\\B = \frac{{{{2016}^{2015}} + 1}}{{{{2016}^{2016}} + 1}} = \frac{{2016.\left( {{{2016}^{2015}} + 1} \right)}}{{2016.\left( {{{2016}^{2016}} + 1} \right)}} = \frac{{{{2016}^{2016}} + 2016}}{{{{2016}^{2017}} + 2016}}\\{2016^{2016}} + 1 < {2016^{2016}} + 2016 \Rightarrow \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2017}^{2017}} + 1}}\\{2017^{2017}} + 1 > {2016^{2017}} + 2016\\ \Rightarrow \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2016}^{2017}} + 2016}}\\ \Rightarrow A < B\end{array}\) Trả lời
Ta có :
A = 2016 ^ 2016 + 1 / 2006 ^ 2007 + 1 < 2006 ^ 2006 + 2005 + 1 / 2006 ^ 2007 + 2005 + 1 = 2006 ^ 2006 + 2006 / 2006 ^ 2007 + 2006 = 2006 . ( 2005 + 1 ) / 2006 . ( 2006 + 1)
= 2006 ^ 2005 +1 / 2006 ^ 2006 + 1 = B
=> A < B
Chúc bn học giỏi nha ! ( sorry , mk viết hơi khó hiểu)
Đáp án:
\[A < B\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}}\\
B = \frac{{{{2016}^{2015}} + 1}}{{{{2016}^{2016}} + 1}} = \frac{{2016.\left( {{{2016}^{2015}} + 1} \right)}}{{2016.\left( {{{2016}^{2016}} + 1} \right)}} = \frac{{{{2016}^{2016}} + 2016}}{{{{2016}^{2017}} + 2016}}\\
{2016^{2016}} + 1 < {2016^{2016}} + 2016 \Rightarrow \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2017}^{2017}} + 1}}\\
{2017^{2017}} + 1 > {2016^{2017}} + 2016\\
\Rightarrow \frac{{{{2016}^{2016}} + 1}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2017}^{2017}} + 1}} < \frac{{{{2016}^{2016}} + 2016}}{{{{2016}^{2017}} + 2016}}\\
\Rightarrow A < B
\end{array}\)