Đáp án:
$\left(-21;-3 ;-\dfrac{1}{3};-\dfrac{1}{3} \right); \left(-1;-1 ;-7;-1\right)$
Giải thích các bước giải:
$\circledast c=0\\\Rightarrow ac=0 \ne 7(L)\\\circledast d=0\\\Rightarrow bd=0 \ne 1(L)\\\circledast c \ne 0; d \ne 0$
$\left\{\begin{array}{l} 3c+a=-22(1)\\ 3d+b=-4(2)\\ad+bc=8(3)\\ac=7(4)\\bd=1(5)\end{array} \right.\\ (1)(4)\Rightarrow \left\{\begin{array}{l} a=\dfrac{7}{c}\\ a=-22-3c\end{array} \right.\\ \Rightarrow \dfrac{7}{c}=-22-3c\\ \Leftrightarrow 7=-22c-3c^2\\ \Leftrightarrow-3c^2 -22c-7=0\\ \Leftrightarrow \left[\begin{array}{l} c=-\dfrac{1}{3} \Rightarrow a=-21 \\ c=-7 \Rightarrow a=-1 \end{array} \right.\\ (2)(5)\Rightarrow \left\{\begin{array}{l} b=\dfrac{1}{d}\\ a=-4-3d\end{array} \right.\\ \Rightarrow \dfrac{1}{d}=-4-3d\\ \Leftrightarrow 1=-4d-3d^2\\ \Leftrightarrow -3d^2-4d-1=0\\ \Leftrightarrow \left[\begin{array}{l} d=-\dfrac{1}{3} \Rightarrow b=-3 \\ d=-1 \Rightarrow b=-1 \end{array} \right.$
Thử vào $(3)$ thấy $(a,b,c,d)=\left(-21;-3 ;-\dfrac{1}{3};-\dfrac{1}{3} \right); \left(-1;-1 ;-7;-1\right)$ thoả mãn.
