Đáp án:
a.$b=c=2012$
b.$x=\dfrac{11}6, y=\dfrac56, z=-\dfrac{13}6$
Giải thích các bước giải:
a.Ta có:
$\dfrac{a}b=\dfrac bc=\dfrac ca=\dfrac{a+b+c}{b+c+a}=1$ vì $a+b+c\ne 0$
$\to a=b, b=c, c=a$
$\to a=b=c$
Mà $a=2012\to b=c=2012$
b.Ta có:
$\dfrac{y+x+1}{x}=\dfrac{x+z+2}{y}=\dfrac{z+y-3}{z}=\dfrac1{x+y+z}=\dfrac{(y+x+1)+(x+z+2)+(z+y-3)}{x+y+z}$
$\to \dfrac{y+x+1}{x}=\dfrac{x+z+2}{y}=\dfrac{z+y-3}{z}=\dfrac1{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}$
$\to \dfrac{y+x+1}{x}=\dfrac{x+z+2}{y}=\dfrac{z+y-3}{z}=\dfrac1{x+y+z}=2$
Vì $\dfrac1{x+y+z}=2\to x+y+z=\dfrac12$
Ta có: $\dfrac{y+x+1}{x}=\dfrac{x+z+2}{y}=\dfrac{z+y-3}{z}=2$
$\to \begin{cases}y=x-1\\ x+z+y+2=3y\\ y=z+3\end{cases}$
$\to \begin{cases}y=x-1\\ \dfrac12+2=3y\\ y=z+3\end{cases}$
$\to \begin{cases}x=y+1\\ \dfrac52=3y\\ z=y-3\end{cases}$
$\to \begin{cases}x=\dfrac{11}6\\ y=\dfrac56\\ z=-\dfrac{13}6\end{cases}$
