Cho a;b;c;d\(e\)0

Biết:\(\dfrac{\text{b+c+d}}{\text{a}}\)=\(\dfrac{\text{c+d+a}}{\text{ }\text{b}}\)=\(\dfrac{\text{d+a+b}}{\text{c}}\)=\(\dfrac{\text{a+b+c}}{\text{d}}\)=\(k\)

Tính\(k\)

Các câu hỏi liên quan