Tính giá trị của biểu thức : A = \(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}+\dfrac{1}{209}+\dfrac{1}{275}\)
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Ta có :
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}\)
\(C=\dfrac{2}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\right)\)
\(C=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(C=\dfrac{2}{3}\left(1-\dfrac{1}{19}\right)\)
\(C=\dfrac{2}{3}.\dfrac{18}{19}=\dfrac{12}{19}\)
~ Học tốt ~
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)
\(=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+\dfrac{2}{10\cdot13}+\dfrac{2}{13\cdot16}+\dfrac{2}{16\cdot19}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{16\cdot19}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{19}\right)=\dfrac{2}{3}\cdot\dfrac{18}{19}=\dfrac{12}{19}\)
Ta có :
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(\Rightarrow C=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{1\cdot4\cdot+4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+\dfrac{3}{16\cdot19}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{18}{19}\)
\(=\dfrac{12}{19}\)
\(C=\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
\(=\dfrac{2}{4}+\dfrac{2}{28}+\dfrac{2}{70}+\dfrac{2}{130}+\dfrac{2}{208}+\dfrac{2}{304}\)
\(=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+\dfrac{2}{13.16}+\dfrac{2}{16.19}\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}.\left(1-\dfrac{1}{19}\right)\)
\(=\dfrac{2}{3}.\dfrac{18}{19}=\dfrac{12}{19}\)
=1/1.2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19=2/1.4+2/4.7+2/7.10+2/10.13+2/13.16+2/16.19(ca tu va mau cung nhan them 2 gia tri khong doi)=2/3.(3/1.4+3/4.7+3/7.10+3/10.13+3/13.16+3/16.19)=2/3.(1/1-1/4+1/4-1/7+1/7-1/10+1/101/13+1/13-1/16+1/16-1/19)=2/3.18/19=12/19
\(\dfrac{2x}{4}+\dfrac{2x}{28}+\dfrac{2x}{70}+...+\dfrac{2x}{418}=\dfrac{14}{11}\)
\(\dfrac{2x}{1.4}+\dfrac{2x}{4.7}+\dfrac{2x}{7.10}+\dfrac{2x}{10.13}+...+\dfrac{2x}{19.22}=\dfrac{14}{11}\)
Ta có công thức : \(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)
Dựa vào công thức ta có :
\(\dfrac{2x}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}\right)=\dfrac{14}{11}\)
\(\dfrac{2x}{3}.\left(1-\dfrac{1}{22}\right)=\dfrac{14}{11}\)
\(\Leftrightarrow\) \(\dfrac{2x}{3}.\left(\dfrac{21}{22}\right)=\dfrac{14}{11}\)
\(\Rightarrow\) \(\dfrac{2x}{3}=\dfrac{14}{11}:\dfrac{21}{22}=\dfrac{4}{3}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Ai thấy đúng thì ủng hộ nha !!!!
1. Tính nhanh:
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}\)
\(=\dfrac{3}{8}\)
2. Tính nhanh
Đặt \(A\) = \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A\) \(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
\(2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(2A=\dfrac{1}{3}-\dfrac{1}{13}\)
\(2A=\dfrac{10}{39}\)
\(A=\dfrac{10}{39}:2\)
\(A=\dfrac{5}{39}\)
\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2\cdot7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}\)
\(\Rightarrow A=-\frac{1}{19}\)