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\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+....+\frac{3^2}{97.100}\)
\(A=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
\(A=3.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(A=3.\left(\frac{1}{1}-\frac{1}{100}\right)=3-\frac{3}{100}=\frac{297}{100}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}+...+\frac{3^2}{97.100}\)
\(A=\frac{3}{1}-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}+\frac{3}{13}-\frac{3}{16}+...+\frac{3}{97}-\frac{3}{100}\)
\(A=\frac{3}{1}-\frac{3}{100}\)
\(A=\frac{297}{100}\)
x/1.4+x/4.7+x/7.10+x/10.13+x/13.16=5/2
=>x/3(1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16)=5/2
=>x/3.(1/4-1/16)=5/2
=>x/3.3/16=5/2
=>x/3=5/2:3/16
=>x/3=40/3
=>x=40
Vậy x=40
x/1.4 + x/4.7 + x/7.10 + x/10.13 + x/13.16 = 5/6
=> x.1/3.(3/1.4 + 3/4.7 + 3/7.10 + 3/10.13 + 3/13.16) = 5/6
=> x.1/3.(1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16) = 5/6
=> x.1/3.(1 - 1/16) = 5/6
=> x.1/3.15/16 = 5/6
=> x.1/3 = 5/6 : 15/16
=> x.1/3 = 8/9
=> x = 8/9 : 1/3
=> x = 8/3
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
a/ \(A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
=> \(A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
=> \(A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
=> \(A=\frac{1}{3}-\frac{1}{9}=\frac{2}{9}\)
b/ \(B=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)
=> \(B=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
=> \(B=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
=> \(B=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{16}\right)=\frac{2}{3}.\frac{15}{16}=\frac{5}{8}\)
\(A=\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+.....+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{97.100}\right)\)
Ta thấy :
\(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
\(.........\)
\(\frac{3}{97.100}=\frac{100-97}{97.100}=\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow A=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)=3\cdot\frac{99}{100}=\frac{297}{100}\)
đáp án = \(\frac{297}{100}\)
đúng không?
kết bạn với mh nha
\(\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+\frac{2}{10\cdot13}+\frac{2}{13\cdot16}\)
\(=2\left(\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}\right)\)
\(=2\left[\frac{1}{3}\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\right)\right]\)
\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\right]\)
\(=2\left[\frac{1}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\right]\)
\(=2\left[\frac{1}{3}\cdot\frac{3}{16}\right]\)
\(=2\cdot\frac{1}{16}\)
\(=\frac{2}{16}=\frac{1}{8}\)
Ta có :
\(\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}\)
\(=\)\(2\left(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=\)\(\frac{2}{3}\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=\)\(\frac{2}{3}.\frac{3}{16}\)
\(=\)\(\frac{1}{8}\)
Chúc bạn học tốt ~
Nguyễn Huy Thắng giải sai rồi ,thế này mới đúng nè
1,\(\frac{1}{6}+\frac{1}{12}+.........+\frac{1}{72}\)
=\(\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{8.9}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{8}-\frac{1}{9}\)
=\(\frac{1}{2}-\frac{1}{9}\)
=\(\frac{7}{18}\)
2,\(\frac{3}{1.4}+\frac{3}{4.7}+..........+\frac{3}{13.16}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{13}-\frac{1}{16}\)
=\(1-\frac{1}{16}\)
=\(\frac{15}{16}\)
2)đặt B= 3/1.4+3/4.7+3/7.10+3/10.13+3/13.16
\(B=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)
\(B=3-\frac{15}{16}\)
\(B=\frac{45}{16}\)
\(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+...+\frac{3^2}{97.100}\)
\(=3.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)
\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3.\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}\)