Question

[Nguyễn Ngọc Trâm]

tính:

a. G= \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)

b. H= \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{8.9.10}\)

m.n làm giúp e bài này vs ạ! e đang cần gấp

m.n trình bày rõ hộ e nha!

camon nhiều ạ <3

Answer 1

[Linh_Windy]

\(linh_1=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{4.5}\right)\)

\(linh_1=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{2}.\dfrac{9}{20}=\dfrac{9}{40}\)

\(linh_2=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{8.9.10}\)

\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\)\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\)

\(linh_2=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{90}\right)=\dfrac{1}{2}.\dfrac{22}{45}=\dfrac{11}{45}\)

Answer 2

[Nguyễn Thanh Hằng]

a/ \(G=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}\)

\(\Leftrightarrow2G=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{1.2}-\dfrac{1}{4.5}\)

\(\Leftrightarrow2G=\dfrac{1}{2}-\dfrac{1}{20}\)

\(\Leftrightarrow2G=\dfrac{9}{20}\)

\(\Leftrightarrow G=\dfrac{9}{40}\)

b/ \(H=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{8.9.10}\)

\(\Leftrightarrow2H=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+.....+\dfrac{2}{8.9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)

\(\Leftrightarrow2H=\dfrac{1}{2}-\dfrac{1}{90}\)

\(\Leftrightarrow2H=\dfrac{22}{45}\)

\(\Leftrightarrow H=\dfrac{22}{90}\)