The basic inequality is super intractable 130, and the basic inequality is super incomprehensible

The requirements for the establishment of the equal sign are too high.

For example, the first step is to require a 2 = 1 (ab) = 1 a (a-b).

The second step requires b=a-b

Three conditions, two unknowns, generally unattainable.

You're using two inequalities.

The first step requires two equal conditions.

a+b+c) 3>=3rd root (ABC).

The equal sign satisfies the condition a=b=c

You didn't think about that.

So. It is useless to be satisfied in the end.

The right way to do this is to find a way to get two equal conditions instead of three.

Minimated: Add AB first, then subtract AB

The reason is. There is ab+1 ab, 1 (a 2-ab) so that there is +a 2-ab

a2+1/（a*b)+1/[a*(a-b)]=ab+1/ab + 1/(a^2-ab)+ a^2-ab)

ab + (1 ab) > = 2 root number 1 = 2, when the equal sign is true, ab = 1 ab, ab = 1, because a>b>0

1 (a 2-ab)]+a 2-ab)>=2, when the equal sign holds, 1 (a 2-ab)=a 2-ab, a 2-ab=1

a 2 = 2, a = root number 2, b = 1 root number 2

So the minimum is 4

The most common mistake of inequality ... It is not possible to derive the maximum value from the preceding equal sign to establish the condition.

The right process is:

Let t=a2+1 ab+1 a(a-b).

Then t=a 2-ab+1 a(a-b)+ab+1 ab 4 equal sign is established as follows.

a^2-ab=1

ab=1 solves a=2b=2

1) (x-a)(x-a^2)<0

There are a and a 2 roots, and you need to consider the size of these two roots!

So when a>a 2, i.e., 01, a0

The root has 2 and (a-2) (a-1).

a>1, (a-2) (a-1) < 2, so x< (a-2) (a-1) or x>2

a=1, x>2

a<1, a-1<0 is [(1-a)x+a-2](x-2)<0 according to the property of inequality 3

When 0a<0, (a-2) (a-1).

1.The solution set of the inequality ax bx c 0 is {x 丨-1 x 2}a<0 (because the solution set of the original inequality is taken to the middle, and the image shows that the free a<0 is in accordance with it).

x=-1 and x=2 are the two roots of the equation ax bx c=0.

From Veddhar's theorem.

x1+x2=1=-b/a

x1*x2=-2=c/a

i.e. b = -a, c = -2a

The inequality a(x 1) b(x-1) c 2ax is equivalent to.

ax²-3ax>0

a<0

x²-3x<0

The solution of the original inequality of 0 a(x 1) b(x-1) c 2ax is {x 丨 0 x 3}

①m=0②m＞0

Less than or equal to 0 (the function is not below the x-axis).

4m 2-8m less than or equal to 0

Because m 0 so 0 in total, 0 "m "2

Cross multiplication.

x-3）（x+1）＜0；

x-3) 0 or (x+1) 0

Solve x 3 or x -1;

When x -1, negative is positive, and it is discarded;

So solve: -1 x 3

Polynomialization is formed using cross multiplication.

x^2-2x-3=(x-3)*(x+1)

Require(x-3)*(x+1)=0

The solution yields x=3 or x=-1

To make (x-3)*(x+1)<0

x has a range of -1

From the original factorization, we get: (x-3)(x+1)<0, and we get -1 by the number line

Let this quadratic equation equal to zero, solve, and then plot, -1

You can find the square of the square that it looks like, and then add a constant to the form of ().