Question

Cho 3 số x y z thỏa mãn:

x²+2yy+1=y²+2z+1=z²+2x+1=0

Tính giá trị BT: P= x²⁰¹⁷+y³+z¹⁹

Answer 1

Ta có : \(x^2+2y+1=y^2+2z+1=z^2+2x+1=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+2y+1=0\left(1\right)\\y^2+2z+1=0\left(2\right)\\z^2+2x+1=0\left(3\right)\end{matrix}\right.\)

Cộng (1) (2) (3) lại ta được :

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

\(\Leftrightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

Do : \(\left(x+1\right)^2\ge0;\left(y+1\right)^2\ge0;\left(z+1\right)^2\ge0\)

\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)

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

Thay vào biểu thức P ta được :

\(P=\left(-1\right)^{2017}+\left(-1\right)^3+\left(-1\right)^9=-3\)