Question

Tìm x,y,z

(t^2+4)(x^2-1)(y^2-9)(z+6)=0

Answer 1

\(\left(t^2+4\right)\left(x^2-1\right)\left(y^2-9\right)\left(z+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t^2+4=0\\x^2-1=0\\y^2-9=0\\z+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t^2=-4\left(voli\right)\\x^2=1\\y^2=9\\z=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t^2=-4\left(voli\right)\\x=\pm1\\y=\pm3\\z=-6\end{matrix}\right.\) 

Vậy khi `x=1` hoặc `x=-1` thì \(t,y,z\in R\), `y=-3` hoặc `y=3` thì \(t,x,z\in R\), `z=-6` thì \(t,x,y\in R\)