The question of the nature of the inequality of the second year of high school, the basic nature of

In a binary linear equation with the variable x a*x 2+b*x+c=0.

b 2-4ac denotes the discriminant formula for the solution of this equation.

At 0, the equation has two solutions; There is no solution to the equation at 0 = There is a solution when 0.

In this problem, you can move the equation to the right of the inequality sign to the left to form an inequality with a variable.

i.e. a 2 + b 2 + ab + 1 - a - b > 0

You can make f(a)=a2+b 2+ab+1-a-b

Let f(a)=0 constitute a binary linear equation with a variable a 2+b 2+ab+1-a-b=0

In this equation =(b-1) 2-4(b 2-b+1)=-3(b-1 3) 2-3 8

You can see that this discriminant is the square of a number multiplied by a negative number plus a negative number, which is definitely less than zero.

i.e. <0

So there's no solution to this equation, so f(a) can't be equal to 0

Because f(a) represents a parabola where a is a variable, a has a coefficient of 1 and the opening is upward.

You can draw it on paper, this parabola has a minimum point, because f(a) is not equal to 0, so the parabola must not have an intersection point with the horizontal axis, so the ordinate of the lowest point of the parabola must be greater than 0. (Otherwise, if the ordinate of the lowest point is less than 0, the left side of the parabola decreases and the right side increases, which must be larger than the coordinates of the lowest point, and must have an intersection point with the horizontal axis).

So f(a)>0 is definitely true.

So f(a)=a 2+(b-1)a+b 2-b+1>0 is definitely true.

Then move a+b to the right of the equation again, so a 2 + b 2 + ab + 1 > a + b

This is found using a quadratic equation of one element, and trigonometric representation of the equation = b 2-4ac

Here a is seen as x and b as a constant.

f(a) is actually a replacement of the original form.

As for Yes. I don't know how to say it, but I want to find if there is a solution to the quadratic inequality.

The coefficient of the primary term is -4x the coefficient of the second term is the constant term.

<> basic property 1: Add (or subtract) the same number (or subtract) on both sides of the inequality and the direction of the inequality sign does not change

If a>ba > b, then a c> b c a c > b c

Basic property 2: The inequality is multiplied (or divided) by the same positive number by two traces, and the direction of the inequality sign is unchanged.

If A>BA > B, and C>0C > 0, then AC>BC AC>BC BC (or AC>BC Frac AC > Frac B C).

Basic property 3: The inequality is multiplied (or divided) by the same negative number on both sides, and the direction of the inequality sign changes

If A>B A > B, and C<0 C < 0, then AC <>

Note For inequalities containing "≠", multiplying (or dividing) a number that is not 0 is still "≠".

If both sides of an inequality are multiplied by 0 at the same time, the inequality becomes an equation.

1.Definition.

The formula that uses an unequal sign to express an inequality relationship is called an inequality The unequal sign encompasses the ">".

For example, 3<4, 2x 50, a 2≠ a+2, etc

1<=a+b<=5===>1/2<=(a+b)/2<=5/2---1)

1<=a-b<=3==>-5/2<=5(a-b)/2<=15/2---2)

1) +2) Got:

1/2-5/2<=3a-2b<=5/2+15/2-2<=3a-2b<=10

Why can't it be counted that way? Do you need linear programming? How simple and how to come)

1) Turn the formula upside down, i.e., prove the reciprocal on the left and the reciprocal on the right;

After taking the reciprocal number, multiply the top and bottom of the right side by (a+b) to make the left and right denominators consistent.

Left denominator = a 2 + b 2, right denominator = (a + b) 2 = a 2 + b 2 + 2ab

a>b>0, so a 2 + b 2 < (a + b) 2, the same as a 2-b 2 unchanged.

Proven. 2) Move all items to the left, proving them "0

With the matching method, 2a 2-4ab+5b 2-2b+1=(2a 2-4ab+2b 2)+3b 2-2b+1

2(a+b)^2+3(b^2-2/3 b+1/9)-1/3 +1

2(a+b)^2+3(b-1/3)^2+2/3

Since the first two terms are perfectly squared, it must be 0, so adding 2 3 must be 2 3, so "0".

Isn't there another condition, about a, b. Otherwise, it's completely discussed.

a²+4b²-4a+4b=a²-4a+4+4b²+4b+1-5(a-2)^2+(2b+1)^2-5

Now let's talk about the size of (a-2) 2+(2b+1) 2-5 and 5.

Then let (a-2) 2+(2b+1) 2-5>5 get the condition, and let (a-2) 2+(2b+1) 2-5<5 get the other condition.

But that doesn't seem to make much sense.

m-n=a²+4b²-4a+4b-5

Then if the formula is greater than 0, then m is greater than n