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dưới đây là phép tính sai ; hãy sửa Thành phép tính đúng ?
123 x 123 = ? ; 5 =? -202099 =?
Bài toán hỏi gì vậy Nguyễn Võ Hoài Thanh
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
\(\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\)
\(=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{7}\)
\(=\dfrac{140}{140}-\dfrac{35}{140}+\dfrac{28}{140}-\dfrac{20}{140}\)
\(=\dfrac{113}{140}\)
| #Sahara |
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{140-35+28-20}{140}=\dfrac{113}{140}\)
Mình nghĩ đề bài cần thêm 1/20 mới tính nhanh được
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(3\times A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(3\times A-A=3-\frac{1}{729}=\frac{2186}{729}\)
\(2\times A=\frac{2186}{729}=>A=\frac{1093}{729}\)
#)Giải :
Bài 1 :
\(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{1280}\right)-\frac{1}{1280}\)
\(\Rightarrow A\times2=\frac{2}{5}-A-\frac{1}{1280}\)
\(\Rightarrow A\times2+A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{2}{5}-\frac{1}{1280}\)
\(\Rightarrow A=\frac{511}{1280}\)
#)Giải :
Bài 2 :
\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{59049}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^9}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{10}}\right)\)
\(2B=1-\frac{1}{3^{10}}\)
\(B=\frac{1-\frac{1}{3^{10}}}{2}\)