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\(=\dfrac{30+20+12+1920+21870}{15+120+960+10935}\simeq1.98\)
Ta có: C = (1.5.6+2.10.12+4.20.24+..+9.45.54):(1.3.5+2.6.10+4.12.20+...+9.27.45)
= (1.3.5.2+2.6.10.2+4.12.20.2+...+9.27.45.2):(1.3.5+2.6.10+4.12.20+...+9.27.45)
= (2.(1.3.5+2.6.10+4.12.20+...+9.27.45)):(1.3.5+2.6.10+4.12.20+...+9.27.45)
=2
\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+9.27.45}\)
\(=\frac{1.5.6+2.1+5.2+6.2+9.1+9.5+9.6}{1.3.5+2.1+2.3+2.5+9.1+9.3+9.5}\)
\(=\frac{1.5.6+\left(1.5.6\right).\left(2.2.2\right)+\left(1.5.6\right).\left(9.9.9\right)}{1.3.5+\left(1.3.5\right).\left(2.2.2\right)+\left(1.3.5\right).\left(9.9.9\right)}\)
\(=\frac{1.5.6.\left(1.2.2.2.9.9.9\right)}{1.3.5.\left(1.2.2.2.9.9.9\right)}=2\)
\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}=2\)
Ta có:
\(C=\frac{1.5.6+2.10.12+4.20.24+...+9.45.54}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=\frac{1.3.5.2+2.6.10.2+4.12.20.2+...+9.27.45.2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=\frac{\left(1.3.5+2.6.10+4.12.20+...+9.27.45\right).2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(=2\)
Vậy C=2.
\(A=\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(A=\frac{1.5.6.\left(1^3+2^3+4^3+9^3\right)}{1.3.5.\left(1^3+2^3+4^3+9^3\right)}=2\)
(1.5.6 + 5.25.30) : (1.3.5 + 5.25.15) = 2
(1.5.6+5.25.30) : (1.3.5+5.25.15)
= (5.6+5.25.30) : (3.5+5.25.15)
= (30+125.30) : (15+125.15)
= (30.1 +125.30) : (15.1+125.15)
= 30. (125+1) : 15(125+1)
= 30.126 : 15.126
= (126:126) . (30:15)
=1.2
=2