\(A=\frac{1.5.6+1.5.6.2^3+1.5.6.3^3+1.5.6.4^3+1.5.6.5^3}{1.3.5+1.3.5.2^3+1.3.5.3^3+1.3.5.4^3+1.3.5.5^3}\)

\(=\frac{1.5.6.\left(1+2^3+3^3+4^3+5^3\right)}{1.3.5.\left(1+2^3+3^3+4^3+5^3\right)}\)

\(=\frac{1.5.6}{1.3.5}=\frac{1.5.3.2}{1.3.5}=2\)

\(=\frac{1.5.6+\left(1.5.6\right).2+\left(1.5.6\right).3+\left(1.5.6\right).4+\left(1.5.6\right).5}{1.3.5+\left(1.3.5\right).2+\left(1.3.5\right).3+\left(1.3.5\right).4+\left(1.3.5\right).5}\)

\(=\frac{\left(1.5.6\right).\left(1+2+3+4+5\right)}{\left(1.3.5\right).\left(1+2+3+4+5\right)}=\frac{1.5.6}{1.3.5}=\frac{1.1.2}{1.1.1}=2\)

A= \(\frac{1.5.3.2+2.10.2.6+2.15.9.2+4.20.12.2+5.25.15.2}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)

A= \(\frac{2+2+2\cdot2+2+2}{0+0+3+0+0}\)

A= \(\frac{12}{3}\)

A= 4

Đầu tiên bạn tách ra, rút gọn rồi cộng lại,tính nha!

bạn ơi mình nhầm, Nguyễn Tiến Dũng đúng, mình làm cách này dài hơn xíu:

=\(\frac{1.5.6+2.10.12+3.15.18+4.20.24+5.25.30}{1.3.5+2.6.10+3.9.15+4.12.20+5.15.25}\)

=\(\frac{6+12+18+24+30}{3+6+9+12+15}\)(*)

= \(\frac{90}{45}\)= \(2\)

Để có đc (*) thì bạn triệt tiêu nha! 

A=\(\dfrac{1.5.6+2.\left(1.5.6\right)+3.\left(1.5.6\right)+4.\left(1.5.6\right)+5.\left(1.5.6\right)}{1.3.5+2.\left(1.3.5\right)+3.\left(1.3.5\right)+4.\left(1.3.5\right)+5.\left(1.3.5\right)}\)

A=\(\dfrac{\left(1+2+3+4+5\right).\left(1.5.6\right)}{\left(1+2+3+4+5\right).\left(1.3.5\right)}\) = \(\dfrac{1.5.6}{1.3.5}\) = 2

bang 2 nhé

\(\frac{1\cdot5\cdot6+2\cdot10\cdot12+4\cdot20\cdot24+9\cdot45\cdot54}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+9\cdot27\cdot45}=\frac{1\cdot5\cdot6\cdot\left(1+2+4+9\right)}{1\cdot3\cdot5\cdot\left(1+2+4+9\right)}=2\)

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

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

Từ đề bài, ta có:

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

Thế này thì đúng rồi, sai sao được :)))

Chúc bạn học tốt!

VÌ MỖI TÍCH TƯƠNG ỨNG TRÊN TỬ SỐ THÌ GẤP 2 LẦN TÍCH Ở MẪU :1.5.6=2.(1.3.5) NÊN TA NHÓM 2 RA

ĐỀ: MK HƠI LƯỜI BN VIẾT ĐỀ RA ĐÂY NHA

\(Â\)\(=\)\(\frac{2.\left(1.5.3+2.10.6+4.20.12+9.45.27\right)}{1.3.5+2.6.10+4.12.20+9.27.45}\)

\(Â\)\(=\)\(2\)

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

\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)=\(\frac{1.5.6+\left(1.5.6\right)2+\left(1.5.6\right)4+\left(1.5.6\right)9}{1.3.5+\left(1.3.5\right)2+\left(1.3.5\right)4+\left(1.3.5\right)9}\)

=\(\frac{\left(1.5.6\right)\left(1+2+4+9\right)}{\left(1.3.5\right)+\left(1+2+4+9\right)}=\frac{1.5.6}{1.3.5}=\frac{6}{3}=2\)

\(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}=\frac{1.5.6+\left(1.5.6\right)2+\left(1.5.6\right)4+\left(1.5.6\right)9}{1.3.5+\left(1.3.5\right)2+\left(1.3.5\right)4+\left(1.3.5\right)9}=\)

\(\frac{\left(1.5.6\right)\left(1+2+4+9\right)}{\left(1.3.5\right)\left(1+2+4+9\right)}=\frac{1.5.6}{1.3.5}=\frac{6}{3}=2\)

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

\(=\frac{1.5.6+\left(1.5.6\right)2+\left(1.5.6\right)4+\left(1.5.6\right)9}{1.3.5+\left(1.3.5\right)2+\left(1.3.5\right)4+\left(1.3.5\right)9}\)

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

\(=\frac{1.5.6}{1.3.5}=\frac{6}{3}=2\)