x？ -x+ 1+2m？ +2mx

=x？ +(2m- 1)x+(2m？ + 1)

=[x+(2m- 1)/2]？ +2m？ + 1-[(2m- 1)/2)？

=[x+(2m- 1)/2]？ +m？ +m+3/4

∵m？ +m+3/4=(m+ 1/2)？ + 1/2 >0

∴[x+(2m- 1)/2]^2+m^2+m+3/4>； 0

∴x？ -x+ 1+2m？ +2mx & gt； 0

∴x？ -x+ 1 & gt； -2m？ -2mx

2. Prove:

( 1)

∫a+b+c = 0 a > b > c

∴a>； 0 degrees celsius & lt0 b-c>；; 0

∴ab-ac=a(b-c)>； 0

Namely AB-AC > 0

∴ab>ac

(2)

∫a+b+c = 0 a > b > c

∴a>； 0 degrees Celsius-LT0 means that A is positive and C is negative.

(1) if b≥0, the minimum value of b is 0, and the maximum value is infinitely close to a, so that 0 = 0/a ≤ b/a.

2 if b

The maximum value of b is infinitely close to 0, when b/a

So I got a B.

To sum up, the range of b/a is:-1/2.

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