Question

8.已知实数x,y满足xy-4x-y+1=0,且x>1,则6x+2y+1的最小值为 ()A. 22？

Answer 1

答案：最小值为27
解析：y（x-1）=4x-1
y=4x-1/(x-1）
6x+2（4x-1）/(x-1）+1
=（8x-8+6）/（x-1）+6（x-1）+7
=6/（x-1）+6（x-1）+15
≥2√36+15=27

Answer 2

由xy-4x-y+1=0得y=(4x-1)/(x-1)=4+3/(x-1),
6x+2y+1=6x+6/(x-1)+9
=6(x-1)+6/(x-1)+15
≥12+15=27,
当x=2,y=7时取等号，
所以6x+2y+1的最小值是27.

Answer 3

解答:xy-4x-y+1=0，y=(4x-1)/(x-1),
设u=6x+2y+1=6x+9+6/(x-1)
u'=6-6/(x-1)²=0，x=2，此时y=7
u最小=14+12+1=27

Answer 4

假设3x+2y=b,那么y=b/2-3x,带入等式得：
(b/2-3x)x-4x-(b/2-3x)+1=0
bx-6x^2-8x-b+6x+2=0
b(x-1)=6x^2+2x-2
b=(6x^2+2x-2)/(x-1)
=(6x^2-12x+6+14x-14+6)/(x-1)
=6(x-1)+6/(x-1)+14
>=2√36+14=26
当(x-1)=1/(x-1)的时候，b取到最小值
(x-1)^2=1,所以x=0（舍去）或x=2
所以b=26,3x+2y+1=27