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1) \(16^{2020}+\dfrac{1}{16^{2021}}+1\)
\(=16^{2021}\div16^{2020}+1\)
\(=16+1\)
\(=17\)
2) \(16^{2021}+\dfrac{1}{16^{2022}}+1\)
\(=16^{2022}\div16^{2021}+1\)
= 17
Vì 17=17 nên \(16^{2020}+\dfrac{1}{16^{2021}}+1=16^{2021}+\dfrac{1}{16^{2022}}+1\)
Gọi phân số đó là \(\dfrac{x}{y}\) .
Theo đề bài ,ta có :
\(\dfrac{x}{y}\)( chuyển 5 đơn vị từ mẫu lên tử thì giá trị \(\dfrac{x}{y}=1\) )
\(\dfrac{x}{y}\) (chuyển 9 đơn vị từ tử xuống mẫu thì giá trị \(\dfrac{x}{y}=\dfrac{3}{5}\) )
Nghĩa là :
\(\left\{{}\begin{matrix}x\\y-5\end{matrix}\right.\) và \(\left\{{}\begin{matrix}x-9\\y\end{matrix}\right.\)
Vậy ta có 2 trường hợp :
Trường hợp 1 : (chuyển 5 đơn vị từ mẫu lên tử)
Ta có :
Ban đầu: \(\dfrac{x}{y}=?\rightarrow\dfrac{x}{y}\) (chuyển 5 đơn vị lên tử) \(\Rightarrow\dfrac{x}{y}=1\)
Vậy :\(\left\{{}\begin{matrix}x\\y-5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x-5\\y-5\end{matrix}\right.\)
Trường hợp 2 : (chuyển 9 đơn vị từ tử xuống mẫu)
Ta có :Ban đầu :\(\dfrac{x}{y}=?\rightarrow\dfrac{x}{y}\)(chuyển 9 đơn vị từ tử)\(\Rightarrow\dfrac{x}{y}=\dfrac{3}{5}\)
\(A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2+...+\left(\dfrac{1}{2013}\right)^2\)
\(A=\left(\dfrac{1}{2+3+4+...+2013}\right)^2\)
\(A=\left(\dfrac{1}{\left(2013-2\right)+1}\right)^2\)
\(A=\left(\dfrac{1}{2012}\right)^2\)
\(A=\dfrac{1}{2012\cdot2012}\)
\(\Rightarrow A=\dfrac{1}{2012}< \dfrac{3}{4}\)
\(\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{25}{8}+...+\dfrac{897}{128}\)
\(=\dfrac{3+9+25+...+897}{2+4+8+...+128}\)
\(=\dfrac{\left(897-3\right)\div6+1}{\left(128-2\right)\div2+1}\)
\(=\dfrac{150}{64}\)
\(=\dfrac{75}{32}\)
\(E=\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{6}-1\right)\cdot\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(E=1\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{45}\right)\)
\(E=1\cdot\left(\dfrac{1}{45}-\dfrac{1}{3}\right)\div2+1\)
\(E=\dfrac{1}{22}\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}\right)\cdot...\cdot\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{101}\right)\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{101}-\dfrac{1}{3}\right)\div2+1\right)\)
\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\dfrac{1}{50}\right)\)
\(P=4^{25}\cdot\dfrac{1}{100}\)
10.
Sửa lại đề :Cho \(P=\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2012}{2013}+\dfrac{2013}{2009}\).Chứng tỏ rằng P<5.
\(P=\dfrac{2009}{2010}+\dfrac{2010}{2011}+\dfrac{2012}{2013}+\dfrac{2013}{2009}\)
\(P=\dfrac{2011}{2012}\)
\(\Rightarrow P< 5\)
\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{200^2}\)
\(A=\dfrac{1}{\left(3+4+5+...+200\right)^2}\)
\(A=\dfrac{1}{\left(200-3+1\right)^2}\)
\(A=\dfrac{1}{198^2}\)
\(A=\dfrac{1\cdot198}{198}\)
\(A=\dfrac{1\cdot2\cdot9\cdot11}{2\cdot3\cdot11}\)
\(A=\dfrac{1\cdot3}{9}\)
\(A=\dfrac{1}{3}\)
Vậy A <1
\(A=\dfrac{1}{1\cdot21}+\dfrac{1}{2\cdot22}+\dfrac{1}{3\cdot23}+...+\dfrac{1}{80\cdot100}\)
\(A=1\cdot\dfrac{1}{21}+\dfrac{1}{2}\cdot\dfrac{1}{22}+...+\dfrac{1}{80}\cdot\dfrac{1}{100}\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{80}\right)-\left(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{100}\right)\)
\(A=1-\left(\dfrac{1}{\left(80-2\right)+1}\right)-\left(\dfrac{1}{\left(100-22\right)+1}\right)\)
\(A=1-\dfrac{1}{79}-\dfrac{1}{79}\)
\(A=1\)
Ta có: A=1.
\(B=\dfrac{1}{1\cdot81}+\dfrac{1}{2\cdot82}+\dfrac{1}{3\cdot83}+...+\dfrac{1}{20\cdot100}\)
\(B=1\cdot\dfrac{1}{81}+\dfrac{1}{2}\cdot\dfrac{1}{82}+\dfrac{1}{3}\cdot\dfrac{1}{83}+...+\dfrac{1}{20}\cdot\dfrac{1}{100}\)
\(B=1-\left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}\right)-\left(\dfrac{1}{81}+\dfrac{1}{82}+\dfrac{1}{83}+...+\dfrac{1}{100}\right)\)
\(B=1-\left(\dfrac{1}{\left(20-1\right)+1}\right)-\left(\dfrac{1}{\left(100-81\right)+1}\right)\)
\(B=1-\dfrac{1}{20}-\dfrac{1}{20}\)
\(B=1\)
\(\Rightarrow\) Ta có tỉ số của \(\dfrac{A}{B}=1\)