Question

tìm x;y;z biết:

\(\left(x+2\right)^2+\left(y-3\right)^4+\left(z-5\right)^6=0\)

Answer 1

\(\Rightarrow\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)

\(\left(y-3\right)^4=0\Rightarrow y-3=0\Rightarrow y=3\)

\(\left(z-5\right)^6=0\Rightarrow z-5=0\Rightarrow z=5\)

Answer 2

Ta có:

\(\left(x+2\right)^2\ge0;\left(y-3\right)^4\ge0;\left(z-5\right)^6\ge0\)

=> để: \(\left(x+2\right)^2+\left(y-3\right)^4+\left(z-5\right)^6=0\)

=>  x + 2 = 0  ; y - 3 = 0  và   z - 5 = 0

=>  x = -2       ;  y = 3     và     z = 5