作业帮初三二元一次方程数学题求解!
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初三二元一次方程数学题求解!
初三二元一次方程数学题求解!
初三二元一次方程数学题求解!
(1)
-3x^2-6x+1
=-3(x^2+2x+1)+3+1
=-3(x+1)^2+4
(2)
(2/3)y^2+(1/3)y-2
=(2/3)[y^2+(1/2)y+(1/16)]-(2/3)*(1/16)-2
=(2/3)[y+(1/4)]^2-(1/24)-2
=(2/3)[y+(1/4)]^2-(49/24)
x^2-2x-3=0
===> (x^2-2x+1)-1-3=0
===> (x-1)^2=4
===> x-1=2,或者x-1=-2
===> x=3,或者x=-1
x^2+(1/x^2)+2[x+(1/x)]=1
===> [x^2+(1/x^2)+2]+2[x+(1/x)]=3
===> [x+(1/x)]^2+2[x+(1/x)]+1=3+1=4
===> [x+(1/x)+1]^2=4
===> x+(1/x)+1=±2
4x^2+y^2-4x+6y+11
=4(x^2-x)+(y^2+6y)+11
=4[x^2-x+(1/4)]+(y^2+6y+9)+1
=4[x-(1/2)]^2+(y+3)^2+1
因为[x-(1/2)]^2≥0,(y+3)^2≥0
所以原式≥1
则它始终为正数
当x=1/2,y=-3时有最小值=1