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\(a^2-\frac{3}{5^2}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)
\(a^2-\frac{3}{5^2}=2.\left(\frac{1}{2.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}+\frac{1}{19.22}+\frac{1}{22.25}\right)\)
\(a^2-\frac{3}{5^2}=2.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{22}-\frac{1}{25}\right)\)
\(a^2-\frac{3}{5^2}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{25}\right)\)
=> \(a^2-\frac{3}{25}=\frac{2}{3}.\frac{23}{50}=\frac{23}{75}\)
=> \(a^2=\frac{23}{75}+\frac{3}{25}=\frac{32}{75}\)
=> \(a=\sqrt{\frac{32}{75}}\)(Nếu thế thì đây phải là đề của lớp 7 chứ nhỉ)
Giải:
\(A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
Gọi: \(B=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\)
\(B=\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}+\dfrac{1}{19.22}+\dfrac{1}{22.25}\right):\dfrac{1}{2}\) \(B=\left[\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{19.22}+\dfrac{3}{22.25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\)
\(B=\left[\dfrac{1}{3}.\dfrac{24}{25}\right]:\dfrac{1}{2}\)
\(B=\dfrac{8}{25}:\dfrac{1}{2}\)
\(B=\dfrac{16}{25}\)
\(\Rightarrow A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\)
\(A^2=\dfrac{16}{25}+\dfrac{9}{25}\)
\(A^2=1\)
\(\Rightarrow A^2=1^2\) hoặc \(A^2=\left(-1\right)^2\)
\(A=1\) hoặc \(A=-1\)
Chúc bạn học tốt!
\(VP=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot5}+\dfrac{1}{5\cdot13}+\dfrac{1}{13\cdot8}+\dfrac{1}{8\cdot19}+\dfrac{1}{19\cdot11}+\dfrac{1}{11\cdot25}\\ =\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}+\dfrac{2}{19\cdot22}+\dfrac{2}{22\cdot25}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}+\dfrac{3}{19\cdot22}+\dfrac{3}{22\cdot25}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\left(1-\dfrac{1}{25}\right)\\ =\dfrac{2}{3}\cdot\dfrac{24}{25}\\ =\dfrac{16}{25}\)\(a^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\\ a^2-\dfrac{9}{25}=\dfrac{16}{25}\\ a^2=\dfrac{16}{25}+\dfrac{9}{25}\\ a^2=1\\ \Rightarrow\left[{}\begin{matrix}a=1\\a=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}-a=-1\\-a=1\end{matrix}\right.\)
Vậy \(-a=-1\) hoặc \(-a=1\)
Cảm ơn bạn Mới Vô đã giúp mình giải bài toán này nha!! Cảm ơn lắm lắm!!!!