Question

已知a>0,实数x,y,z满足x+y+z=a,x2+y2+z2=a2/2,求x的取值范围

Answer 1

由x+y+z=a得y+z=a-x;平方得y^2+z^2+2yz=x^2+a^2-2ax,x2+y2+z2=a2/2两式相减得

Yz=a^2/4+x^2-ax,则y、z为t^2-(a-x)t+( a^2+x^2-ax)=0两根,则Δ=( a-x)^2-4(a^2/4+x^2-ax)≥0整理得-3x^2+2ax≥0所以0≤x≤2/3a