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\(\dfrac{5}{2}A=\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{9-4}{4.9}+\dfrac{14-9}{9.14}+\dfrac{19-14}{14.19}+...+\dfrac{509-504}{504.509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\)
\(\dfrac{5}{2}A=\dfrac{1}{4}-\dfrac{1}{509}\)
\(A=\left(\dfrac{1}{4}-\dfrac{1}{509}\right).\dfrac{2}{5}\)
\(A=\dfrac{1}{10}-\dfrac{2}{2545}< \dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{1}{10}\)(đpcm)
Chúc bạn học tốt!
Ta có:
A=\(\dfrac{1}{2.9}+\dfrac{1}{9.7}+\dfrac{1}{7.19}+...+\dfrac{1}{252.509}\)
A=2.(\(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{504.509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{504}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{1}{4}-\dfrac{1}{509}\))
A=\(\dfrac{2}{5}\).(\(\dfrac{509}{2036}-\dfrac{4}{2036}\))
A=\(\dfrac{2}{5}\).\(\dfrac{505}{2036}\)
A=\(\dfrac{101}{1018}\)
b)\(\dfrac{1}{7}B=\dfrac{1}{10.18}+\dfrac{1}{18.26}+\dfrac{1}{26.34}+...+\dfrac{1}{802.810}\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{8}{10.18}+\dfrac{8}{18.26}+\dfrac{8}{26.34}+...+\dfrac{8}{802.810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{34}+...+\dfrac{1}{802}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}\left(\dfrac{1}{10}-\dfrac{1}{810}\right)\)
\(\dfrac{1}{7}B=\dfrac{1}{8}.\dfrac{8}{81}\)
\(\dfrac{1}{7}B=\dfrac{1.8}{8.81}\)
\(\dfrac{1}{7}B=\dfrac{1}{81}\)
\(B=\dfrac{1}{81}:\dfrac{1}{7}\)
\(B=\dfrac{7}{81}\)
Bài toán tổng quát: Đề này n lẻ mới đúng nhé
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{ac+bc+c^2}\right)=0\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{ab\left(ac+bc+c^2\right)}=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Nếu \(a=-b\Rightarrow a^n=-b^n\) và \(\dfrac{1}{a^n}=\dfrac{-1}{b^n}\)
Ta có: \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{c^n}\)
\(\dfrac{1}{a^n+b^n+c^n}=\dfrac{1}{c^n}\)
VT = VP => ĐPCM
Còn ý còn lại thì dựa trên bài này mà biến đổi một tí là ra
a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)
b: 
A=1/2.9+1/9.7+1/7.19+...+1/252.509
=?
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