Question

已知4x-3y-6z=0,x+2y-7z=0,则x²+5y²+7z²分之2xy+2yz-6

Answer 1

4x - 3y -6z = 0 .(1)式

x + 2y - 7z = 0 .(2)式

4 * (2)式 - (1)式

11y - 22z = 0 => y = 2z

带入(2)式得到 x = 7z - 2y = 3z

所以所求的式子

= ( (3z)² + 5 * (2z)² + 7z²) / ( 2* 3z * 2z + 2* 2z * z - 6* 3z *z)

= [(9 + 20 + 7)z²] / [(12 + 4 - 18) z²]

= 36 / -2

= -18