Let a = 2x & # 178; - x + 1, B = x & # 178; + X, then the size relation of a and B

Let a = 2x & # 178; - x + 1, B = x & # 178; + X, then the size relation of a and B

a≥b
a=2x²-x+1
b=x²+x
a-b=(2x²-x+1)-(x²+x)
a-b=2x²-x+1-x²-x
a-b=x²-2x+1
a-b=(x-1)²
Because: (x-1) &# 178; ≥ 0
So: A-B ≥ 0
Therefore: a ≥ B

It is used to compare the size of a & # 178; - A + 1 and 3 / 4
Is there any other way, because I remember another way is to get a & # 178; - A + 1 / 4 and then use the discriminant (B & # 178; - 4ac) to judge

First make a difference: a2-a + 1-3 / 4 = a2-a + 1 / 4 = (A-1 / 2) 2, because (A-1 / 2) 2 ＞ = 0 constant holds, so a2-a + 1 constant ＞ = 3 / 4, if you don't understand, you can continue to ask questions

Compare the difference: (x + 4) & #, (x + 2) (x + 6)

（x＋4）²-（x＋2）（x＋6）
=x²+8x+16-(x²+8x+12)
=4>0
So (x + 4) & # > (x + 2) (x + 6)