High school math problems about inequalities, hurry! 5

It is better to discuss this question in the way of combining numbers and shapes.

m is obviously not equal to 0

f(x)=2mx^2-2(4-m)x+1

Axis of symmetry x = (4-m) 2m

Maximum: [8m-4*(4-m) 2)] (8m).

1 m>0, f(x)=2mx 2-2(4-m)x+1 open up, just draw.

For g(x)=mx, x>0, g(x)>0, x=0, g(x)=0, in this case, f(0)>0 i.e. f(0)=1

x<0,f(x)>0,1) axis of symmetry is to the right of the y-axis, i.e., (4-m) 2m>0, i.e., m<4So: 0=1

2) The symmetrical car is on the left side of the y-axis, i.e., (4-m) 2m<0,m>4, and the minimum value: [8m-4*(4-m) 2)] (8m)<>0

When x tends to infinity, both f(x) and g(x) will be negative!

In summary: 0

When all 3 values <=0, then x:x<=0, g(x):m>=0, f(x): m<=0, (8-2m) 2-4*2m*1<0

So, there is no solution.

So there is at least one positive number, and m belongs to r

Well, it doesn't seem right, that's the answer. But I don't know how to write it, I don't know if I'm right.

Yes 9. The process of finding is that the condition is set by the question, a> ridge 2, b>3, b 3=a (a-2)=1+2 (a-2).

2a+b/3=2a+1+2/(a-2)=2[(a-2)+1/(a-2)]+5。Application of the Fundamental Inequality, 2a+b 3 9. Where, a=3, b=9, "= holds.

Let u=2 a, v=3 b, then u,v>0, u+v=1, a=2 u, b=3 v, y=2a+b 3=4 u+1 v=4 u+1 (1-u).

4-3u) (u-u 2) with x=4 3-u, then u=4 3-x, y=3x [4 3-x-(4 3-x) 2].

3x/(-4/9+5x/3-x^2)

4 (9x)+x 4 3, when x=2 Tongsen 3, u=1 3, take the equal sign, so y 3 (5 3-4 guess mill 3) = 9, so the smallest value of y is 9

x-x-2 0 gives x<-1 or x>2

2x +(2k+5)x+5k=(x+k)(2x+5) 0 gives -k-5 2, and since there is only an integer solution of -2, x<-1 is solved in front

So -2<-k -1

So in summary, 1 k<2

x -x-2=(x-2)(x+1) 0 --x<-1 , or x>2 is the set of integer solutions, so x<-1 is taken

2x²+（2k+5)x+5k=(2x+5)(x+k)<0 --5/2k<2

The value range of the real number k is k<2

Arithmetic Mean = Geometric Mean.

x+8 x>=2 (root number 8).

f(x)=16x/(x²+8)

16 (x+8 x)<=16 2 (root number 8) = 2 (root number 2) b -3b 21 4

b 2-3b-(3 2) 2+21 4-(3 2) 2=(b-3 2) 2+3>=3>2 (root number 2)>=f(a).

The first question is that the numerator and denominator are the same as the x numerator, which is x+8 x, you know.

If a>0, then b -4ac 0, the inequality ax +bx+c 0 is constant because a>0, so f(x)=ax +bx+c the opening of the function goes up b -4ac 0, so the function has only one intersection point with the x-axis, so the function is not less than zero In addition, you can also formulate.

ax +bx+c=a(x+b 2a) -b -4ac) 4a can draw the conclusion you want.

Binary once?How can there be a duality?

Hehe, just know:

When a>0 and b -4ac 0, ax +bx+c 0 is held in constant.

It is extremely doubtful that this question is about the unary quadratic inequality [ax 2+bx+c 0](a>0), which holds constant at b 2-4ac<0.

If the title is as I said, then:

Let y=ax 2+bx+c

When b 2-4ac<0, the equation has no solution.

That is, the image y=ax 2+bx+c does not intersect with the x-axis.

Since a>0, the image opening is upward.

So, AX 2+BX+C 0 is always established.

When a is less than zero, the parabola opening is downward, there is no intersection with the axis x, and the image is all below the y-axis, so it is always true.

Isn't it squared?Yes, it should be less than zero, because when it is less than zero, there is no intersection with the x column, and it will be greater than zero in the definition domain r, otherwise there is an intersection with x, and it is not Evergrande in r. It will be well understood, and you will get twice the result with half the effort!

ax 2+bx+c 0 Evergrande is 0

According to the y=ax 2+bx+c image, the opening direction is upward, and there can only be one intersection point with the x-axis.

ax+bx+c≥0

Is this a binary inequality? There's only one unknown.

Because a, b, and c are not equal, so:

LG(A+B) 2+LG(B+C) 2+LG(C+A) 2>LG AB+LG Shed Slip BC+LG CA=LG(ABC).

and LGA+LGB+LGC=LG(ABC).

lg(a+b)/2+lg(b+c)/2+lg(c+a)/2>lga+lgb+lgc

Wild 2)((n+1) 2+1)-(n+1)=1 ( n+1) 2+1)+(n+1)).

n^2+1)-n=1/(√n^2+1)+n)((n+1)^2+1)+(n+1)>√n^2+1)+n>01/(√n+1)^2+1)+(n+1))<1/(√n^2+1)+n)>√n+1)^2+1)-(n+1)<√n^2+1)-n1+1/2^+1/3^+.1/n^2

1+1/(1*2)+1/(2*3)+.1 (n*(n-1))1+1 (1*2)+1 Song and Harmony (2*3)+.1/(n*(n-1))1+1-1/2+1/2-1/3+..

1/(n-1)-1/n2-1/n<2

1+1/2^+1/3^+.1/n^2<2

a^a*b^b/(a^b*b^a)=(a^a/a^b)*(b^b/b^a)=a^(a-b)*b^(b-a)

a b) (a-b), when a>b>0, a b>1, a-b>0, (a b) (a-b)> a b) 0=1, so a a*b b>(a b*b a).

When b>a>0, 0(a b) 0=1, so a a*b b>(a b*b a).

In summary: a a*b b> (a b*b a).

This question should first be discussed on a case-by-case basis! What does it mean that the symbol does not change? It's always positive or negative! Then the steps to solve the problem will come out naturally!

I do it according to the idea of combining numbers and shapes!

First of all, when f(x) > 0, just make f(-2) >0 and f(2) >0! (because it is a one-time function, it is monotonic), and the solution is <