Question

Tìm x,y,z

\(\left(x-1\right)^2+\left(3x-y-3\right)^2+\left(y+z\right)^4=0\)

Answer 1

\(\left(x-1\right)^2+\left(3x-y-3\right)^2+\left(y+z\right)^4=0\)

\(\left(x-1\right)^2\ge0\)

\(\left(3x-y-3\right)^2\ge0\)

\(\left(y+z\right)^4\ge0\)

\(\left(x-1\right)^2+\left(3x-y-3\right)^2+\left(y+z\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^2=0;\left(3x-y-3\right)^2=0;\left(y+z\right)^4=0\)

\(x-1=0\Rightarrow x=1\)\(3x-3-y=0\Rightarrow3\times1-3=y\Rightarrow y=0\)\(y+z=0\Rightarrow0+z=0\Rightarrow z=0\)

Vậy \(x=1;y=0;z=0\)

Chúc bạn học tốt ^^