cho x,y ∈ N thỏa mãn : (3x+5y)(x+4y) chia het cho 7. Cmr (3x+5y)(x+4y)chia het cho 49 29/09/2021 Bởi Melody cho x,y ∈ N thỏa mãn : (3x+5y)(x+4y) chia het cho 7. Cmr (3x+5y)(x+4y)chia het cho 49
$\begin{array}{l}\text{- Ta có : $(3x+5y)(x+4y)\ \vdots\ 7$}\\\text{mà 7 là số nguyên tố}\\\to\left[\begin{array}{l} 3x+5y\ \vdots\ 7\\x+4y\ \vdots\ 7\end{array}\right.\\\text{+ Xét $3x+5\ \vdots\ 7$}\\\to2(3x+5y)\ \vdots\ 7\\\to 6x+10y\ \vdots\ 7\\\to 7(x+2y)-(6x+10y)\ \vdots\ 7\\\to 7x+14y-6x-10y\ \vdots\ 7\\\to x+4y\ \vdots\ 7\\\to (3x+5y)(x+4y)\ \vdots\ 7^2\\\to (3x+5)(x+4y)\ \vdots\ 49\quad(1)\\\text{+ Xét $x+4y\ \vdots\ 7$}\\\to 10(x+4y)\ \vdots\ 7\\\to 10x+40y\ \vdots\ 7\\\to (10x+40y)-7(x+5y)\ \vdots\ 7\\\to 10x+40y-7x-35y\ \vdots\ 7\\\to 3x+5y\ \vdots\ 7\\\to (3x+5y)(x+4y)\ \vdots\ 7^2\\\to (3x+5y)(x+4y)\ \vdots\ 49\quad(2)\\\text{- Từ (1) và (2) $\to (3x+5y)(x+4y)\ \vdots\ 49$ (đpcm)} \end{array}$ Bình luận
$\begin{array}{l}\text{- Ta có : $(3x+5y)(x+4y)\ \vdots\ 7$}\\\text{mà 7 là số nguyên tố}\\\to\left[\begin{array}{l} 3x+5y\ \vdots\ 7\\x+4y\ \vdots\ 7\end{array}\right.\\\text{+ Xét $3x+5\ \vdots\ 7$}\\\to2(3x+5y)\ \vdots\ 7\\\to 6x+10y\ \vdots\ 7\\\to 7(x+2y)-(6x+10y)\ \vdots\ 7\\\to 7x+14y-6x-10y\ \vdots\ 7\\\to x+4y\ \vdots\ 7\\\to (3x+5y)(x+4y)\ \vdots\ 7^2\\\to (3x+5)(x+4y)\ \vdots\ 49\quad(1)\\\text{+ Xét $x+4y\ \vdots\ 7$}\\\to 10(x+4y)\ \vdots\ 7\\\to 10x+40y\ \vdots\ 7\\\to (10x+40y)-7(x+5y)\ \vdots\ 7\\\to 10x+40y-7x-35y\ \vdots\ 7\\\to 3x+5y\ \vdots\ 7\\\to (3x+5y)(x+4y)\ \vdots\ 7^2\\\to (3x+5y)(x+4y)\ \vdots\ 49\quad(2)\\\text{- Từ (1) và (2) $\to (3x+5y)(x+4y)\ \vdots\ 49$ (đpcm)} \end{array}$