\(A=\frac{3x5x7+6x10x14+9x15x21}{5x7x9+10x14x18+15x21x27}\)
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\(A=\frac{2.6.10+6.10.14+...+1994.1998.2002}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{1.2^3.3.5+3.2^3.5.7+....+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=8\)
Vậy A = 8
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\(A=9\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\)
\(=9\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\)
\(=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)=\dfrac{260}{87}\)
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\(\frac{4}{1x3x5}+\frac{4}{3x5x7}+...+\frac{4}{9x11x13}\)
\(=\frac{1}{1x3}-\frac{1}{3x5}+\frac{1}{3x5}-...+\frac{1}{9x11}-\frac{1}{11x13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
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\(\dfrac{4}{1\times3\times5}+\dfrac{4}{3\times5\times7}+\dfrac{4}{5\times7\times9}+\dfrac{4}{7\times9\times11}\)
=\(\dfrac{5-1}{1\times3\times5}+\dfrac{7-3}{3\times5\times7}+\dfrac{9-5}{5\times7\times9}+\dfrac{11-7}{7\times9\times11}\)
=\(\dfrac{1}{1\times3}-\dfrac{1}{3\times5}+\dfrac{1}{3\times5}-\dfrac{1}{5\times7}+...+\dfrac{1}{9\times11}-\dfrac{1}{11\times13}\)
=\(\dfrac{1}{3}-\dfrac{1}{143}=\dfrac{140}{429}\)
\(A=\frac{1}{3}\)
\(\frac{3.5.7+6.10.14+9.15.21}{5.7.9+10.14.18+15.21.27}\)
= \(\frac{3.5.7+3.2.5.2.7.2+3.3.5.3.7.21}{5.7.9+5.2.7.2.9.2+5.3.7.3.9.3}\)
= \(\frac{3.5.7+3.5.7.2.2.2+3.5.7.3.3.3}{5.7.9+5.7.9.2.2.2+5.7.9.3.3.3}\)
= \(\frac{3.5.7+3.5.7.8+3.5.7.27}{5.7.9+5.7.9.8+5.7.9.27}\)
= \(\frac{3.5.7.\left(1+8+27\right)}{5.7.9.\left(1+8+27\right)}\)
= \(\frac{3.5.7.36}{5.7.9.36}\)
= \(\frac{1}{3}\)