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a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)
a/b= 7/6 + 7/10 + 7/12
a/b= 7(1/6+1/10+1/12)
Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 - \(\frac{1}{2018}\)= \(\frac{2017}{2018}\)< 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2018^2}\)< 1 ( dpcm )
Ta có:
\(\frac{1}{2^2}\)< \(\frac{1}{1.2}\).
\(\frac{1}{3^2}\)< \(\frac{1}{2.3}\).
\(\frac{1}{4^2}\)< \(\frac{1}{3.4}\).
...
\(\frac{1}{2017^2}\)< \(\frac{1}{2016.2017}\).
\(\frac{1}{2018^2}\)< \(\frac{1}{2017.2018}\).
Từ trên ta có:
\(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+...+ \(\frac{1}{2016.2017}\)+ \(\frac{1}{2017.2018}\)= 1- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{2016}\)- \(\frac{1}{2017}\)+ \(\frac{1}{2017}\)- \(\frac{1}{2018}\)= 1- \(\frac{1}{2018}\)< 1.
=> \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+...+ \(\frac{1}{2017^2}\)+ \(\frac{1}{2018^2}\)< 1.
=> ĐPCM.
Giải thích thêm: ta thấy \(\frac{1}{2^2}>\frac{1}{100}\),...,\(\frac{1}{10^2}=\frac{1}{100}\)=> từ \(\frac{1}{2^2}\)đến \(\frac{1}{10^2}\)có 5 cặp
\(\frac{1}{12^2}< \frac{1}{100}\),...,\(\frac{1}{100^2}< \frac{1}{100}\)=> từ \(\frac{1}{12^2}\)đến \(\frac{1}{100^2}\)có 45 cặp
=> 45>5 => tổng < 1/2 (kết hợp với cái kia nx thì bn mới hiểu)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{24.25}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
\(=\frac{1}{5}-\frac{1}{25}\)
\(=\frac{4}{25}\)
b, \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
Gọi biểu thức trên là A
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
\(A=1+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}\)
\(A=1+2^2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+..+\frac{1}{2009^2}\right)\)
Ta có: \(\frac{1}{3^2}< \frac{1}{1.3};\frac{1}{5^2}< \frac{1}{3.5};\frac{1}{7^2}< \frac{1}{5.7};...;\frac{1}{2009^2}< \frac{1}{2007.2009}\)
\(\Rightarrow A< 1+4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{2007.2009}\right)\)
\(=1+4\cdot\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=1+2\left(1-\frac{1}{2009}\right)=3-\frac{2}{2009}< 3\)
\(\Rightarrow A< 3\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
Ta có:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}:2=\frac{25}{51}\)