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.

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

Bằng 2 bạn nhé

k cho mk dc k.đi..................................................

\(ĐặtA=\frac{1.5.6+2.10.12+4.20.24+...+9.45.54}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)

\(A=\frac{1.5.6\left(1+2.2.2+4.4.4+...+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+.....+9.9.9\right)}\)

\(A=\frac{1.5.6}{1.3.5}\)Vậy ta có \(A=2\)

\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(=\frac{1.5.6+2^3.1.5.6+4^3.1.5.6+9^3.1.5.6}{1.3.5+2^3.1.3.5+4^3.1.3.5+9^3.1.3.5}\)

\(=\frac{1.5.6\left(1+2^3+4^3+9^3\right)}{1.3.5\left(1+2^3+4^3+9^3\right)}=\frac{5.6}{3.5}=2\)

bang 2 nhé

thử khử phân số này là 1 tính rồi

\(A=\frac{1.5.6+2^3.1.5.6+4^3.1.5.6+9^3.1.5.6}{1.3.5+2^3.1.3.5+4^3.1.3.5+9^3.1.3.5}=\frac{1.5.6.\left(1+2^3+4^3+9^3\right)}{1.3.5.\left(1+2^3+4^3+9^3\right)}=2\)