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\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}\)
Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{4};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{1009^2}< \dfrac{1}{1008.1009}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{1009^2}< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1008.1009}\)\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{1009}\)
\(\Rightarrow A< \dfrac{3}{4}\left(đpcm\right)\)
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
A = \(\dfrac{\left(\dfrac{47}{15}+\dfrac{3}{15}\right):\dfrac{5}{2}}{\left(\dfrac{38}{7}-\dfrac{9}{4}\right):\dfrac{267}{56}}=\dfrac{\dfrac{10}{3}.\dfrac{2}{5}}{\dfrac{89}{28}.\dfrac{56}{267}}=2\)
B= \(\dfrac{1,2:\left(\dfrac{6}{5}.\dfrac{5}{4}\right)}{0,32+\dfrac{2}{25}}=\dfrac{\dfrac{6}{5}:\dfrac{3}{2}}{\dfrac{8}{25}+\dfrac{2}{25}}=\dfrac{4}{\dfrac{5}{\dfrac{2}{5}}}=2\)
=> A = B
Bài 1:
Ta có:
\(\left(100a+3b+1\right)\left(2^a+10a+b\right)=225\left(1\right)\)
Mà \(225\) lẻ nên \(\left\{{}\begin{matrix}100a+3b+1\\2^a+10a+b\end{matrix}\right.\) cùng lẻ \(\left(2\right)\)
\(*)\) Với \(a=0\) ta có:
Từ \(\left(1\right)\Leftrightarrow\left(100.0+3b+1\right)\left(2^a+10.0+b\right)=225\)
\(\Leftrightarrow\left(3b+1\right)\left(1+b\right)=225=3^2.5^2\)
Do \(3b+1\div3\) dư \(1\) và \(3b+1>1+b\)
Nên \(\left(3b+1\right)\left(1+b\right)=25.9\) \(\Rightarrow\left\{{}\begin{matrix}3b+1=25\\1+b=9\end{matrix}\right.\) \(\Leftrightarrow b=8\)
\(*)\) Với \(a\ne0\left(a\in N\right)\) ta có:
Khi đó \(100a\) chẵn, từ \(\left(2\right)\Rightarrow3b+1\) lẻ \(\Rightarrow b\) chẵn
\(\Rightarrow2^a+10a+b\) chẵn, trái với \(\left(2\right)\) nên \(b\in\varnothing\)
Vậy \(\left\{{}\begin{matrix}a=0\\b=8\end{matrix}\right.\)
Bài 2:
Ta có:
\(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+...+2017}\)
\(=\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}+\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}+...+\dfrac{1}{\dfrac{\left(1+2017\right).1009}{2}}\)
\(=\dfrac{2}{2.4}+\dfrac{2}{3.6}+\dfrac{2}{4.8}+...+\dfrac{2}{1009.2018}\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{1009.1009}\)
\(\Rightarrow A< \dfrac{1}{2.2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1008.1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1008}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{1009}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\) (Đpcm)
Có \(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...+\dfrac{1}{1+3+5+...+2017}\)
\(\Rightarrow A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{1+3+...+2017}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2017^2}\)
Ta thấy:
\(\dfrac{1}{2^2}=\dfrac{1}{4}\)
\(\dfrac{1}{3^2}< \dfrac{1}{3.2}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
.................
\(\dfrac{1}{2017^2}< \dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2016.2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2016}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}-\dfrac{1}{2017}\)
\(\Rightarrow A< \dfrac{3}{4}\)
Vậy \(A< \dfrac{3}{4}\).
Có \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) +...+ \(\dfrac{1}{1+3+...+2017}\)
= \(\dfrac{1}{2^2 }\)+\(\dfrac{1}{3^2}\) + ... +\(\dfrac{1}{2017^2}\)
Lại có :
\(\dfrac{1}{2^2}\) = \(\dfrac{1}{4} \)
\(\dfrac{1}{3^2}\) <\(\dfrac{1}{2.3}\)
...
\(\dfrac{1}{2017^2}\) <\(\dfrac{1}{2016.2017}\)
\(\Rightarrow \) A< \(\dfrac{1}{4} \) +\(\dfrac{1}{2.3}\)+... +\(\dfrac{1}{2016.2017}\)
A<\(\dfrac{1}{4} \)+\(\dfrac{1}{2}\)- \(\dfrac{1}{3}\) +...+\(\dfrac{1}{2016}- \dfrac{1}{2017}\)
A< \(\dfrac{1}{4} \)+\(\dfrac{1}{2}\) -\(\dfrac{1}{2017}\)
A<\(\dfrac{3}{4}\) -\(\dfrac{1}{2017}\)
\(\Rightarrow\)A<\(\dfrac{3}{4}\) (đpcm)
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