4 high school sophomore inequality solving questions, 4 high school inequality questions,

1. ax bx c 0, divide by x to get c x +b x+a 0 so that t=1 x, you can get ct +bt+a 0

Because 0

So 1 t 1, i.e. the solution set of ct + bt + a 0 is 1 t 1

The solution set of ct +bt+a 0 is t 1 , t 1 so the solution set of cx +bx + a<0 is x 1 , x 1 2, ax 2a 1 x 2 0

Factoring yields (x-2)(ax-1) 0

1 A 2, then 2 x 1 A

0 1 a 2, then 1 a x 2

1 a=2, there is no solution.

1 A 0, then x 1 A, x 2

3、. f(x)＝ax²－2x＋2

When a 0, f(1) 0, f(4) 0

Substitution yields a 3 8

When a 0, the axis of symmetry x=1 a 1 or x=1 a 4 makes f(1) 0, f(4) 0

It can be solved a 1

In summary, the above two situations are discussed.

4、 f(x)＝x²＋ax＋3

a [ 1,1] always has f(x) 0

Because =a -12,a [ 1,1].

So in the range of a [1,1].

A -12 is constant less than 0

In this case, f(x) is in the range r

1, easy to know a<0 , and c<0 makes ax bx c=0 know x1= x2= x1+x2=-b a x1*x2=c a

The solution set of cx +bx+a<0 is x>1 or x<1 2, =4a 2+4>0 so a>0 has no solution, a<0 x r, a=0, x<-2

I'm at work, so I'm not going to do it, let's do two more downstairs.

1.Because the equation has a negative root and no positive root, there is when xamp; lt;At 0, there is a |x|=;Substituting it into the equation is to repent, and the boy pretends to be -x=ax+1So we get x=-1 (a+1)amp; lt;0 then there is aamp; gt;-1;nbsp;When xamp; gt;At 0, there is a |x|=x is substituted into the equation to solve x=ax+1,x=1 (1-a)amp; gt;0 then there is aamp; lt;1.

Because the equation has no positive roots, the range of values of a should be taken as its complement, i.e., aamp; gt;=;So there is aamp; gt;=;2,nbsp;Let a=k -2k+3 2amp; gt;=1/2.When aamp; gt;1, there is a function y=a x is a monotonic increasing function, so by a xamp; lt;a (1-x) we can get xamp; lt;1-x get xamp; lt;1/2.The conditions are met.

When 1 2amp; lt;aamp;lt;1, there is a Bizao function y=a x is monotonically reduced, so by a xamp; lt;a (1-x) we can get xamp; gt;1-x, solve xamp; gt;1 2 Do not meet the conditions. nbsp;When a=1, the condition is obviously not satisfied. nbsp;In summary, there are aamp; gt;1, so we can solve kamp; gt;(2+2) 2 or kamp; lt;(2-√2)/2nbsp;3. Let these two numbers be x, and y is 1 x + 9 y = 1

and (x+y)(1 x+9 y)amp; gt;=(x* 1x+ y*9 y) 2=16, the condition for the equal sign to be true is y=3xSo there is x=4, y=; 4. Deform the inequality to (x-1) 2amp; gt;k^;So there is x-1amp; gt;k2 or xamp; lt;x-1amp;lt;-k^2nbsp;So we can solve the xamp; gt;k 2 + 1 or xamp lt;1-k^2.

Question 1 analysis: To require the minimum value of a+b+c, first of all, a +2ab+4bc+2ac=12 is converted into a certain form of a+b+c, and it is easy to think of the form of (a+b+c) 2 by observing the equation.

From (b-c) 2>=0, b2-c 2>=2bca 2+2ab+4bc+2ac=12

a^2+2ab+2bc+2ac+b^2+c^2>=12(a+b+c)^2>=12

Because a, b, c>0

a+b+c>=2 root 3

The second problem analysis: for the function y=1 x image can be drawn according to the tracing method, the function y=x-1 x can be regarded as the superposition of y=x and y=1 x two functions, this problem does not need to draw the function image, by the function interval 00, the maximum value is x=2 to obtain 3 2

1 solution: a square 2ab + 2ac + 4bc = 12 and: 2bc < = b square + c squared.

Therefore, the original form can be reduced to.

A squared 2AB+2AC+2BC+2BC=12A squared 2AB+2AC+2BC+B squared+C squared》 = 12 (A + B + C) squared "" = 12

a b c>0

a+b+c>=2, root number 3

Solution 2: No, there is a maximum value, and the maximum value is 3 2

When x increases, 1 x decreases, and (-1 x) increases again, so x+(-1 x) also increases.

That is, the value of x-1 x increases with the increase of x, so when x=2 there is a maximum value of 2-1 2=3 2

(a) (a+b+c) = a +b +c +2ab+2bc+2ca12=a +2ab+4bc+2caSubtract the two formulas to give (a+b+c) -12=(b-c) 0

=>(a+b+c)²≥12.===>a+b+c≥2√3.∴(a+b+c)min=2√3.

2) From the definition of monotonicity, it can be seen that when x 0, the function f(x)=x-(1 x) increases, on (0,2], the function f(x) increases, and at x (0,2], there is always f(x) f(2)=3 2That is, when 0 x 2, there is always x-(1 x) 3 2At this point, x-(1 x) has a maximum value of 3 2

x+y)(y+z)=xy+xz+y^2+yz

y(x+y+z)+xz>=2xyz(x+y+z)=2 under the root number

So the minimum value of the original formula is 2

x+y 2 xy x=y, take the minimum value, 2 xyy+z 2, take the minimum value when yz=z, take the minimum value 2, take the minimum value (x+y)(y+z) when yzx=y=z, 4 xy 2z, substitute x=y=z into xyz(x+y+z)=1, and get x=y=z=4 times 1 3 minimum value: 4 xy 2z=4 1 3=4 3 3

1.By the title, sina = 1 3, tanb = 3--- sinb = 3 2, cosc = 3 4--- sinc = 7 4

3/2=√12/4>√7/4---b>c√7/4=√63/12>√16/12=1/3---c>a∴b>c>a

2.From the title, there is: logb(1 b)0

by logb(1b)loga(b)<1

by loga(1b)loga(b)>0

a, b>1, then there are a>b>1

a, b<1, then there is 0b>1 or 01 c-a 4-1 4a-c 5

--1≤c≤7---2≤2c≤14---2≤8a-2c≤10

--6≤9a-3c≤9

--4≤9a-c≤23

--f(3)∈[4,23]

4.Less conditions, right?

When x-1 >0, it is x>1

1<(x-1)(x+1)=x^2-1

x^2>2

x> 2 or x<-2

Because x>1, x> 2 under the root number

When x-1 >0, it is x>1

1>(x-1)(x+1)=x^2-1

x^2<2

2 under the root number because x<1

Therefore, the value of 2x under the root number is the union of the two cases.

1/(x-1)1

When x-1 0, i.e., x 1, the original equation becomes 1 (x+1) (x-1), so -1 x 1, the synthesis is -1 x 1

So the solution set is: x -1 and x ≠1

｜x-2/x｜＞x-2/x

Since the form is the same, it can be judged.

x-2/x<0

x^2-2)/x<0

x+√2)(x-√2)x<0

You'll take care of the rest.

Negative infinity, - 2) and (2, 0).

Understand a>=0 then a = a ; If a<0 then a = -a positive >> a (negative).

So (x-2) x >(x-2) x means (x-2) x <0=> x-2) with x variant sign =>x(x-2)<0 =>0

The upstairs solution is correct x-2 x<0

The answer is either (negative infinity, negative root 2) or (0, positive root, 2).

x+2≥0x^2-4)^2≤(x+2)^2x+2)^2(x-2)^2≤(x+2)^2x+2)^2(x^2-4x+3)≤0

1) When x+2=0, lead hail x=-2, which is eligible;

2) When x+2≠0, Huaifan x 2-4x+3 01 x 3, so the solution of the original inequality is 1 x Yu Hao 3 or x=-2