Inequalities with parameters in the solution 500 points 100

Your question is too complicated, and after the question is done, I will tell you the most basic 2 methods.

1.Variable separation, find the maximum or minimum value of the separated function. (Usually it is necessary to find the derivative many times, and after the first derivation, a known zero point must be required, but because the first zero point is made up, it is not possible to determine whether there are other zero points, so the second derivative is required, because the purpose of the derivative function is to determine the zero point and the sign, so the main purpose of the second derivative is to judge the symbol, if the symbol can be determined, such as: .)

If it is greater than 0, we can determine that the one-derivative function increases monotonically, and we can determine that the zero point just now is a unique zero point. If the symbol cannot be determined for the second time, and only a known zero point can be found, then the derivative is found again by analogy. Then the reverse goes up, and you can find the extreme value of the separated function.

2.Direct derivation. Same method as above.

Principles to be followed in solving inequalities with parameters.

Inequalities with parameters have been one of the key contents of the college entrance examination over the years. When solving inequalities with parameters, due to the uncertainty of the parameters, it is often necessary to comprehensively classify the parameters according to the range of values of the parameters. In problem solving, if we follow the principle of "whether the system is right or not, and then judge the size of the root", the classification criteria are often clear and clear, and there are rules to follow.

This principle is: first, to see whether the coefficient of the highest order of the inequality is positive; Secondly, compare the roots of the corresponding equations, which is larger and which is smaller.

Forget it, I recommend a book to you Dragon Gate Special Topic (Inequality) This book is very good, basically the inequality is very hateful, read it for yourself, I wish you all the best in your studies.

All I can say is, good luck.

Too much put. I can't finish talking about it for a while.

Are you from Hangzhou, I have a way.

Figure out the plus/minus and zeros of the parameters. Classify.

First, take the argument as a constant and make x2-(3-x)x+3c<0x2-3x+x2+3c<0

2x2-3x+3c<0

Equation 2 (x-root number 6 2) squared + 3c-3 2<0 (x-root number 6 2) squared < (3c-3 2) 2 and then discuss c c can only be greater than 1 2 to make sense.

And then talk about x, when x is greater than the root number 6 2 the result of x comes out, notice that the result of x here has to be brought back to the previous x is greater than the root number 6 2, and then reanalyze c.

When x takes the root number 6 2....

When x is less than the root number 6 2. When solving this kind of problem, first analyze the value of the parameter, and then go to the result of x, bring back the range of the parameter, and solve it o( o....

In addition to c, it needs to be discussed in a categorical manner: when c is greater than 0; Less than 0; Equal to 0.

The rest of the sum equations are the same.

1．42x^2+ax-a^2>0

6x+a)(7x-a)>0

If a=0 then the inequality is 42x 2>0

x is not equal to 0 if a>0

then x>a 7, x<-4 6

If a<0

then x>-a 6, x0

If the discriminant formula is less than 0

a^2-8<0

a=-2 2, x is not equal to 2

a=-2 2, x is not equal to 2

If the discriminant formula is greater than 0

a^2-8〉0

a<-2√2,a>2√2

then x>[-a+ (a 2-8)] 2,x<[-a- (a 2-8)] 2

x-1)(ax-1)>0

Discussion of A in five cases.

1, a<0, then 1 a<1

1 a1 x < 1 or x>1 a

4. a=1, then x≠1

5. a>1, then 1 a<1

x<1 a or x>roll-to-smart 1

Added: The classification is based on the relationship between 1 A and 1 and the plus or minus of A.

a（x-1）/x-2＞1

ax-a-x+2 x-2 0 is equivalent to:

ax-x-a+2)*（x-2）＞0

a-1）x+2-a]*（x-2）＞0

When a=1, x (2, +00).

x (-00, a-2 a-1) (2, +00) when a 1, x (2, a-2 a-1) when 0 a 1

Move 1 to the left, divide it, and then discuss A.

（ax-a-x+2）/(x-2)>0

a-1)x-(a-2)] (x-2)>0 is equivalent to:

a-1)x-(a-2)](x-2)>0 When a=1, the above equation becomes x-2>0, when a≠1, the above equation corresponds to two equations as x1=(a-2) (a-1) or x2=2 when x1>x2, that is: when (solve the inequality about a, it is also a fractional inequality, get: 01, when a<0, a-1<0, the solution set is.

I seem to have discussed it a lot.,The landlord deleted it himself.。 Take the part of a>0.

Move the term first to get a(x-1) (x-2)-1>0, and then divide (a-1)x-(a-2) (x-2)>0, sort out [(a-1)x-(a-2)](x-2)> compare the size of (a-2) (a-1) and 2, when (a-2) (a-1) > 2, that is, 01 has x>2 or x

1.When a=1, the original formula is x-1 x-2>1

The shift is reduced to 2 x-2>0

Solve x>2

2.When a is not equal to 1, the original formula is (a-1)x-a+2 x-2>0, that is, the numerator and denominator have the same sign.

1) x>2 and (a-1) x-a+2>0

Solution x>a-2 a-1

Then we will also discuss the size of 2 and a-2 a-1. It's annoying. I just finished the college entrance examination, and I did too much, and I was disgusting to watch.

2) x<2 and (a-1) x-a+2<0

In the same way as solving x, we should also discuss the size of 2 and a-2 a-1.

Give me a point. It's not easy for me, thank you

The above is a bit of a problem, x 2+ax+a<0 is not an equation, it is an inequality and the constant is the maximum value <0

The first floor is wrong, he thinks that x 2 + ax + a = 0 has a solution to give a point idea in [1,4] In fact, there is a solution in [1,4], that is, the minimum value < 0 and the axis of symmetry is -a 2

Adang-a 2<1

In [1,4] for the increase.

The minimum values are f(1)<0

1+a+a<0

24 is minus in [1,4].

The minimum values are f(4)<0

16+4a+a<0

a<-8

c 1<-a/2<4

f(-a/2)<0-8

Let the function y=x 2+ax+a, then the minimum value of the function on [1,4] is less than 0, which means that x 2+ax+a<0 has a solution on [1,4].

There is no need to discuss it by classification, because the minimum value must be taken at the endpoint, and the endpoint has three, that is, the value of -a 2 is less than 0, and the relationship between the three is taken at the time or end, and finally the union is taken.

No! Since x 2+ax+a<0 has a solution on [1,4], then x 2+ax+a=0 has a solution or two solutions at [1,4] f(x)=x 2+ax+a

There is a solution: f(0)*f(4)<0

i.e. a*(16+4a+a)<0

160 3 axes of symmetry within [1,4].

1 Delta 0 gets A>4 or A<0

2 f(4)*f(0)>0 gives a>0 or a<-163 the axis of symmetry is -8 in [1,4], so a range is -16