问2道高一数学题

1.
x^2-4x+1<0

2.两式相减：
x^2+y^2-[2(2x-y)-5]
=x^2+y^2-4x+2y+5
=(x^2-4x+4)+(y^2+2y+1)
=(x-2)^2+(y+1)^2
≥0
两个平方式的和不小于0，且只有当x=2且y=-1时才＝0
所以x^2+y^2≥[2(2x-y)-5]

(x-2-√3)(x-2+√3)<0
(x-2)^2-3<0
x^2-4x+4-3<0
x^2-4x+1<0

x^2+y^2-[2(2x-y)-5]
=x^2+y^2-[4x-2y-5]
=x^2+y^2-4x+2y+5
=x^2-4x+4+y^2+2y+1
=(x-2)^2+(y+1)^2

因为(x-2)^2>=0,(y+1)^2>=0
所以(x-2)^2+(y+1)^2>=0
即x^2+y^2>=2(2x-y)-5

1.
(x - (2-√3) ) ( x - (2+√3) ) =0
x^2 -4x - 1 = 0

2.
x^2+y^2 - (2(2x-y)-5 )
=(x^2 -4x + 4) + (y^2 + 2y + 1)
= ( x-2)^2 + (y+1)^2 ≥ 0

x^2+y^2 ≥ 2(2x-y)-5