Cho hình chữ nhật ABCD có chiều dài AB chiều rộng BC. Nối A với C, trên AC lấy E sao cho AE = 2 EC. Nối E với B và D. a)So sanh diện tích hai tam giác ABE và EDC b)Cho diện tích tam giác EDC = 4,5cm ²Tính độ dài các cạnh hình chữ nhật ABCD?
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\(\dfrac{1}{4}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{70}\)+\(\dfrac{1}{130}\)+\(\dfrac{1}{208}\)+\(\dfrac{1}{304}\)
= \(\dfrac{1}{1x4}\)+\(\dfrac{1}{4x7}\)+\(\dfrac{1}{7x10}\)+\(\dfrac{1}{10x13}\)+\(\dfrac{1}{13x16}\)+\(\dfrac{1}{16x19}\)
= \(\dfrac{1}{3}\)x(\(\dfrac{3}{1x4}\)+\(\dfrac{3}{4x7}\)+\(\dfrac{3}{7x10}\)+\(\dfrac{3}{10x13}\)+\(\dfrac{3}{13x16}\)+\(\dfrac{3}{16x19}\))
= \(\dfrac{1}{3}\)x(\(\dfrac{4-1}{1x4}\)+\(\dfrac{7-4}{4x7}\)+\(\dfrac{10-7}{7x10}\)+\(\dfrac{13-10}{10x13}\)+\(\dfrac{16-13}{13x16}\)+\(\dfrac{19-16}{16x19}\))
= \(\dfrac{1}{3}\)x(\(\dfrac{4}{1x4}\)-\(\dfrac{1}{1x4}\)+\(\dfrac{7}{4x7}\)-\(\dfrac{4}{4x7}\)+\(\dfrac{10}{7x10}\)-\(\dfrac{7}{7x10}\)+\(\dfrac{13}{10x13}\)-\(\dfrac{10}{10x13}\)+\(\dfrac{16}{13x16}\)- \(\dfrac{13}{13x16}\)+ \(\dfrac{19}{16x19}\)-\(\dfrac{16}{16x19}\))
= \(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{13}\)+\(\dfrac{1}{13}\)-\(\dfrac{1}{16}\)+\(\dfrac{1}{16}\)-\(\dfrac{1}{19}\))
= \(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{19}\))
= \(\dfrac{1}{3}\)x\(\dfrac{18}{19}\)
= \(\dfrac{6}{19}\)