If it is the case of ab c, the inequality is constant

Take 2*b=a+c, know n<=4, and prove that 1 (a-b)+1 (b-c)>=4 (a-c) is constant.

Multiply both sides of the inequality by a-c at the same time, and combine a-c = a-b + b-c to get :

2+(a-b) (b-c)+(b-c) (a-b)>=4, where a-b>=0 and b-c>=0 ②

From the mean inequality: (a-b) (b-c)+(b-c) (a-b)>=2, holds, and thus holds.

In summary, the maximum value of n is 4

1 a-b+1 b-c)(a-b+b-c)>=4 (using mean inequality) so (1 a-b+1 b-c)>=4 a-c, so the maximum value of n is 4

Solution: A is true: Reason: If a b, then both sides of the inequality are subtracted by a number, and the sign of the inequality does not change.

b is not established: reason: a b, a b, add c on both sides to get c a c b

c is not true: Reason: Both sides of the inequality are multiplied by a negative number, and the sign direction of the inequality changes.

d is not true: Reason: Both sides of the inequality are divided by a negative number, and the sign direction of the inequality changes.

b -4ac 0 then ax +bx + c>0 is not necessarily true.

It is not sufficient because it is a.

And ax +bx+c>0 will always set up Bu Zai, then b -4ac 0 must be established to return to the brigade.

So it is necessary.

Therefore, it is necessary to not fully strip the world.

a＞0，b＞0

So a+b>0

Take A+B on both sides

Original = 2+a b+b a+k 0

2+ab+ba-k is a mean inequality.

Because. a b+b pat zen a 2 root number a b*b a = 2 so. 4≥-k

k 4, so the minimum value of Naga sedan k is Dong 4

Solution: Answer: D.

Because a>b, c<0Add or subtract a number (positive or negative) on both sides of the inequality at the same time, and the sign of the inequality does not change. So a is correct.

And because both sides of the inequality are multiplied by a negative number at the same time, the inequality has to be changed, so b is correct. a-c must be greater than b, because subtracting a negative number is equivalent to adding its absolute value (negative negative is positive), so c holds. d is not entirely wrong, because if 2>1, but 2+(>1 still holds.

But it's not absolutely true. So this answer can only be D.

Solution: A is true: Reason: If a b, then both sides of the inequality are subtracted by a number, and the sign of the inequality does not change.

b is not established: reason: a b, a b, add c on both sides to get c a c b

c is not true: Reason: Both sides of the inequality are multiplied by a negative number, and the sign direction of the inequality changes.

d is not true: Reason: Both sides of the inequality are divided by a negative number, and the sign direction of the inequality changes.

A is not true, because B may be negative, and only Huai has |a|>|b|to get a 2 > b 2

b is not true, because when c < 0, the lead on both sides of a > b is multiplied by a negative number, and the direction of the unequal sign changes.

c does not hold, because AC 2=BC 2d holds when the bond is c = 0 (according to the nature of inequality).

d for option a, a2

b22ab, so option A is wrong;

For options b and c, although ab > 0, it can only mean that a and b have the same sign, if a and b are less than 0, options b and c are wrong;

For option d, ab>0, <

0, then <>

Therefore, D. was chosen