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\(\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}\)
\(=\dfrac{1.2.3\left(1+2.2.2+4.4.4+7.7.7\right)}{1.3.6\left(1+2.2.2+4.4.4+7.7.7\right)}\)
\(=\dfrac{1.2.3}{1.3.6}=\dfrac{1}{3}\)
Ta có:
\(\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.41}\)
=\(\dfrac{1.2.3\left(1+2^3+4^3+7^3\right)}{1.3.6\left(1+2^3+4^3+7^3\right)}\)= \(\dfrac{1.2.3}{1.3.6}\) = \(\dfrac{1}{3}\)
\(B=\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}=\frac{1.2.3\left(1+2+4+7\right)}{1.3.6\left(1+2+4+7\right)}=\frac{1.2.3}{1.3.6}\frac{2}{6}=\frac{1}{3}\)
Ta có:
\(A=\dfrac{2002}{2000.16-1970}=\dfrac{2002}{\left(2002-2\right).16-1970}\)
\(=\dfrac{2001}{2002.16-2.16-2002}=\dfrac{2002}{2002.16-2002}\)
\(=\dfrac{2002}{2002.\left(16-1\right)}=\dfrac{1}{16-1}=\dfrac{1}{15}\)
\(\Rightarrow A=\dfrac{1}{15}\)
\(B=\dfrac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}\)
\(=\dfrac{1.2.3.\left(1+2.2.2+4.4.4+7.7.7\right)}{1.3.6.\left(1+2.2.2+4.4.4+7.7.7\right)}\)
\(=\dfrac{1.2.3}{1.3.6}=\dfrac{1}{3}\)
\(\Rightarrow B=\dfrac{1}{3}\)
Vì \(\dfrac{1}{3}>\dfrac{1}{15}\Rightarrow A< B\)
Vậy A < B
\(A=\dfrac{2002}{2000.16-1970}=\dfrac{2002}{2002.16-32-1970}=\dfrac{2002}{2002.15}=\dfrac{1}{15}\)
\(B=\dfrac{1.2.3\left(1.1.1+2.2.2+4.4.4+7.7.7\right)}{1.3.6\left(1.1.1+2.2.2+4.4.4+7.7.7\right)}=\dfrac{1}{3}=\dfrac{5}{15}>\dfrac{1}{15}=A\\ \Rightarrow B>A\)
A = 2002 2000.16 − 1970 = 2002 ( 2002 − 2 ) .16 − 1970 = 2002 2002. ( 16 − 1 ) = 1 15
và B = 12.3 + 2.4.6 + 4.8.12 + 7.14.21 1.3.6 + 2.6.12 + 4.12.24 + 7.21.42 = 12.3 + 2.4.6 + 4.8.12 + 7.14.21 1.3.2.3 + 2.6.3.4 + 4.12.3.8 + 7.21.14.3 = 12.3 + 2.4.6 + 4.8.12 + 7.14.21 3. ( 12.3 + 2.4.6 + 4.8.12 + 7.14.21 ) = 1 3 = 5 15
Vậy B > A
\(\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.6+2.6.12+4.12.24+7.21.42}=\frac{1.2.3\left(1+2+4+7\right)}{1.3.6\left(1+2+4+7\right)}=\frac{1.2.3}{1.3.6}=\frac{2}{6}=\frac{1}{3}\)