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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.
Giá trị của biểu thức \(\frac{1.5.6+2.10.12+4.20.24+...+9.45.54}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(ĐặtA=\frac{1.5.6+2.10.12+4.20.24+...+9.45.54}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(A=\frac{1.5.6\left(1+2.2.2+4.4.4+...+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+.....+9.9.9\right)}\)
\(A=\frac{1.5.6}{1.3.5}\)Vậy ta có \(A=2\)
\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}=2\)
A=\(\frac{1.5.6+8\left(1.5.6\right)+64\left(1.5.6\right)+...+729\left(1.5.6\right)}{1.3.5+8\left(1.3.5\right)+64\left(1.3.5\right)+...+729\left(1.3.5\right)}\)
=\(\frac{\left(1.5.6\right)\left(1+8+64+...+729\right)}{\left(1.3.5\right)\left(1+8+64+...+729\right)}\)
=\(\frac{1.5.6}{1.3.5}\)
=2
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
Bằng 2 nhé