ab = -6 (1)

bc = -15 (2)

ca = 10 (3)

Từ (1) => \(a=-\frac{6}{b}\) .Thay vào (3) ta được: \(c.\left(-\frac{6}{b}\right)=10\Rightarrow c=10:\left(-\frac{6}{b}\right)=-\frac{5}{3}b\)

Thay \(c=-\frac{5}{3}b\) vào (2) ta được: \(b.\left(-\frac{5}{3}b\right)=-15\Rightarrow-\frac{5}{3}b^2=-15\Rightarrow b^2=9\Rightarrow\orbr{\begin{cases}b=3\\b=-3\end{cases}}\)

+ Với b = 3 => \(c=\left(-\frac{5}{3}\right).3=-5\) và \(a=-\frac{6}{3}=-2\)

+ Với b = -3 \(\Rightarrow c=\left(-\frac{5}{3}\right).\left(-3\right)=5\) và \(a=\frac{-6}{-3}=2\)

 Vậy \(\orbr{\begin{cases}a=-2,b=3,c=-5\\a=2,b=-3,c=5\end{cases}}\)

a= 2

b= -3

c=5

ab,cd - a,bcd = 17,865

a,bcd x 10 - a,bcd = 17,865

a,bcd x 9 = 17,865

a,bcd = 17,865 : 9

a,bcd = 1,985

a=1;b=9;c=8;d=5

a) Xét ΔABC có

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(Định lí tổng ba góc trong một tam giác)

Ta có: \(\widehat{A}:\widehat{B}:\widehat{C}=6:2:1\)

nên \(\dfrac{\widehat{A}}{6}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{1}\)

mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(cmt)

nên \(\dfrac{\widehat{A}}{6}=\dfrac{\widehat{B}}{2}=\dfrac{\widehat{C}}{1}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{6+2+1}=\dfrac{180^0}{9}=20^0\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{\widehat{A}}{6}=20^0\\\dfrac{\widehat{B}}{2}=20^0\\\dfrac{\widehat{C}}{1}=20^0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{A}=120^0\\\widehat{B}=40^0\\\widehat{C}=20^0\end{matrix}\right.\)

Vậy: \(\widehat{A}=120^0\); \(\widehat{B}=40^0\); \(\widehat{C}=20^0\)