Question

1.x²+x-1=0，则x³+2x²+3=

2.代数式x²+9y²-2x+12y-6的最小值是？，此时x²-y²=？

Answer 1

x³+2x²+3
=x(x²+x-1)+x²+x+3
=x²+x-1+4
=4
x²+9y²-2x+12y-6
=(x²-2x+1)+(9y²+12y+4)-11
=(x-1)²+(3y+2)²-11≥-11
x=1,y=-2/3时 ,代数式x²+9y²-2x+12y-6的最小值是-11
x²-y²=1-4/9=5/9

Answer 2

1，因为x^2+x-1=0
x^3+x^2-x+x^2+x+3
=x(x^2+x-1)+(x^2+x-1)+4
=x*0+x+4
=4
所以x^3+2x^2+3的值是4
2，x^2+9y^2-2x+12y-6
=(x^2-2x+1)+(9y^2+12y+4)-1-4-6
=(x-1)^2+(3y-2)^2-11
因为(x-1)^2+(3y-2)^2>=0
所以(x-1)^2+(3y-2)^2-11>=-11
所以所求代数式的值是-11