giúp tớ
\(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{97.100}\)
\(=3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=3\left(1-\dfrac{1}{100}\right)\)
\(=3.\dfrac{99}{100}=\dfrac{297}{100}\)
Vậy...
\(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{97.100}\)
\(=3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=3\left(1-\dfrac{1}{100}\right)=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(A=\dfrac{3^2}{1\cdot4}+\dfrac{3^2}{4\cdot7}+\dfrac{3^2}{7\cdot10}+\dfrac{3^2}{10\cdot13}+\dfrac{3^2}{13\cdot16}+...+\dfrac{3^2}{97\cdot100}\)
\(A=3\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}+...+\dfrac{3}{97\cdot100}\right)\)
\(A=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3\left(1-\dfrac{1}{100}\right)\)
\(A=3\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)
\(A=3\cdot\dfrac{99}{100}\)
\(A=\dfrac{297}{100}\)
A=3.(3/1.4+3/4.7+3/7.10+....+3/97.100)
A=3.(1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)
A=3.(1-1/100)
A=3.99/100
A=279/100
Vậy A=279/100
Ta có
A= 32/4.7+32/7.10+....+32/97.100
A=3.(3/4.7+3/7.10+....+3/97.100
A=3.(1/4 - 1/7+1/7-1/10+....+1/97-1/100)
A=3.(1/4-1/100)
A=3.6/25
A=18/25
ok \(\dfrac{18}{25}\)
!!!!! .Hãy giúp tớ giải bài ở dưới nhé các bạn yêu quý.
