∵z=7-x-y
∴x²+y²+(7-x-y)²=17
∴x²+y²-7x-7y+xy+16=0
化成关于y的一元二次方程得:y²+(x-7)y+x²-7x+16=0
∵方程有实数解
∴⊿=(x-7)²-4(x²-7x+16)≥0
即3x²-14x+15≤0
∴5/3≤x≤3
x=3+3cost,y=3sint z=x2+y2+2x =(3+3cost)^2+(3sint)^2+2(3+3cost) =24+24cost 0<=z<=48 z=x2+y2+2xμ?è??μ·??§0<=z<=48
x∈[5/3, 3]