Question

Cho x^2 + y^2 + z^2 = 6 và xy - 3x + 2z = 10 
tính P= (1006xy - 2019y - x^3 + z^5) / (x^2 + 2y^3) 

Answer 1

Ta có x² + y² + z² = 6 ; xy - 3x + 2z = 10

Khi đó 4(x² + y² + z²) - 4(xy - 3x + 2z) = 24 - 40

<=> 4x² + 4y² + 4z² - 4xy + 12x - 8z + 16 = 0

<=> (x² - 4xy + 4y²) + (3x² + 12x + 12) + (4z² - 8z + 4) = 0

<=> (x - 2y)² + 3(x + 2)² + 4(z - 1)² = 0

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

Thay x = -2 ; y = -1 ; z = 1 vào P ta được \(P=\dfrac{1006xy-2019y-x^3+z^5}{x^2+2y^3}\)

\(=\dfrac{1006.(-2).(-1)-2019.(-1)-(-2)^3+1^5}{(-2)^2+2.1^3}\)

\(=\dfrac{2020}{3}\)