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Ta có: ( x + 2)( x - 5) = -12
=> \(x+2\inƯ\left(-12\right);x-5\inƯ\left(-12\right)\)
mà Ư (-12) = \(\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x+2\in\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\x-5\in\left\{"....."\right\}\end{matrix}\right.\)
Xét các t/h:
Đặt A = \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
\(\Rightarrow2A=\)\(=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(\Rightarrow2A=\) \(\dfrac{1}{1}-\dfrac{1}{15}=\dfrac{14}{15}\)
\(\Rightarrow A=\dfrac{14}{15}:2=\dfrac{7}{15}\)
Ngan Nguyen
Câu hỏi của Đặng Quý Dương - Toán lớp 6 - Học toán với OnlineMath
\(B=\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)
\(7B=\dfrac{7}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)
\(7B=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)
\(7B=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)
\(7B=\dfrac{1}{2}-\dfrac{1}{28}\)
\(7B=\dfrac{13}{28}\)
\(B=\dfrac{13}{4}\)
câu hỏi ở đâu thanh phuong Do???
yes