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\(D=2!^2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}\right)\)
tổng trong ngoặc nhỏ hơn 1 nên D nhỏ hơn 4.1=4<6
Vậy Đ<6
Ta có: \(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(=2\left(2+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{2}{2015}\)
Vậy D < 6.
\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\)
=>\(D=\frac{\left(1.2\right)^2}{1^2}+\frac{\left(1.2\right)^2}{3^2}+\frac{\left(1.2\right)^2}{5^2}+\frac{\left(1.2\right)^2}{7^2}+...+\frac{\left(1.2\right)^2}{2015^2}\)
=>\(D=\frac{2^2}{1^2}+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2015^2}\)
=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)\)
Ta có: \(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}< 2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)
=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
Mà \(2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)\(=2\left(2+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=2\left(2+1-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{6}{2016}< 6\)
=>\(D< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)< 6\)
=>D<6
\(A=\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6+\frac{7}{4}+\frac{3}{2}\right)\)
\(A=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6-\frac{7}{4}-\frac{3}{2}\)
\(A=\left(3-5-6\right)-\left(\frac{1}{4}+\frac{7}{4}+\frac{3}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{6}{5}\)
\(A=-8-\left(2+\frac{3}{2}\right)+1+\frac{6}{5}\)
\(A=-8-2-\frac{3}{2}+1+\frac{6}{5}\)
\(A=-9-\frac{3}{2}+\frac{6}{5}\)
\(A=\frac{-93}{10}\)
Mk lm đc 1 cách thui
Ủng hộ mk nha ^_-