Question

第5/6题怎么解答，求解，谢谢

Answer 1

5.f(x)=(x+m)^2-(m+3)/2,x>0时f(x)>0分两种情况：
1）m<0,-m>0,f(x)min=f(-m)=-(m+3)/2>0,解得m<-3.
2)m≥0,f(x)在(-m,+∞)是增函数，f(x)>f(0)=m^2-m/2-3/2≥0,解得m≥3/2.
综上，m<-3或m≥3/2.
6.2x-3y=z,
x+3y=14z,
解得x=5z,y=3z,
x≠0,故z≠0,
所以(x^3y+5xyz+xz)/(y^2+z^2)

=(375z^2+75z+5)/10
=(75z^2+15z+1)/2
=(3/2)(25z^2+5z+1/3)
=(3/2)[(5z+1/2)^2+1/12]
当z=-1/10时它取最小值1/8.

Answer 2

5.f(x)=(x+m)^2-(m+3)/2,x>0时f(x)>0分两种情况：
1）m<0,-m>0,f(x)min=f(-m)=-(m+3)/2>0,解得m<-3.
2)m≥0,f(x)在(-m,+∞)是增函数，f(x)>f(0)=m^2-m/2-3/2≥0,解得m≥3/2.
综上，m<-3或m≥3/2.
6.2x-3y=z,
x+3y=14z,
解得x=5z,y=3z,
x≠0,故z≠0,
所以(x^3y+5xyz+xz)/(y^2+z^2)

=(375z^2+75z+5)/10
=(75z^2+15z+1)/2
=(3/2)(25z^2+5z+1/3)
=(3/2)[(5z+1/2)^2+1/12]
当z=-1/10时它取最小值1/8.