When x belongs to 1,2, the inequality x x mx 4 0 is constant, and the value range of m is found

=b -4ac=m -16>0, m>4 or m<-4x*x+mx+4 0

Let the inequality be x + mx + 4 = 0

That is, the equation has two roots, which are x1=(-b+ b -4ac) 2a=(-m+ m -16) 2, x2=(-b- b -4ac) 2a=(-m- m -16) 2, m -16 0, and m 4 or m -4 is obtained;

x1-x2=(-m+√m²-16)/2-(-m-√m²-16)/2=√m²-16≥0

x1≥x2x∈(1,2)

x1>2,x2<1

m+√m²-16)/2>2

m²-16>(4+m)≥0

i.e. m -16> (4+m).

4+m)≥0

m -4, m <-4 (rounded).

m-√m²-16)/2<1

m+ m -16>-2, m -16>-2-m 0 i.e. m -16>-2-m

2-m≥0m<-5,m≤2

In summary, m>4

The opening is upward, and x*x+mx+4 is used to <0

The following conditions must be met: >0

f(1)<0

f(2)<0

Then figure it out for yourself.

Because x*x+mx+4 0 is constant, so m=-(x*x+4) x so m=--[x+4 x] is less than or equal to --2 multiplied by the root number [x*(4 x)] basic inequality) =--4

Let f(x)=x 2+mx+4

then from the known, obtained.

Only f(1) 0 and f(2) 0 are needed to draw the function image, so 1+m+4 0 is m -5

And 4+2m+4 0 is m -4

So, m -5

x is not 0 and is greater than 0

then m then m<(x+2 x)min

Let y=x+2 x

y'=1-2/x^2

Lingy'=0 then x = - root number 2 (return to Qiyan House) or side sail x = root number 2ymin = 2 root number 2

m<2 root number 2< x "Leaky 2< P>

Discussion: x -2 is definitely worth it: -(x+1)-(x+2) m gets -(2x+3) m, because (2x+3) is the increase function of the pants, and -(2x+3) is a subtraction function on the pure sock sense field, so -(2x+3) at x=-2 Qingbi obtains the minimum value (can be drawn) so m 1-2 m -1-(x+1)+(x+2) mm 1 when x -1(x+1)+(x+2) m gets.

x 2+mx+4<0 is constant at x (1,2).

m<-(x 2+4) x=-(x+4 x)-(x+4 x) in the range of x (1,2) is (-5,-4) so m can be fetched to -5, because -(x+4 x) cannot get -5, and m is always less than -(x+4 x), -5 is invariably less than -(x+4 x) (x (1,2)).

One drawing: you know that f(x)=x 2+mx+4 must pass the point (0,4) To f(x)<0 is constant, the axis of symmetry needs to be on the right side of the origin, that is.

m<0 and f(1) 0, f(2) 0, so that the two root >0 are comprehensively solved to obtain the intersection m -5

i.e. m (-5].

Because x (1,2), where x belongs to the open range, m has to go to the closed range.

The second point to mention, in the case of inaccuracy, you can substitute m=-5 into the original equation to see if it meets the meaning of the question, don't you understand?

=b -4ac=m -16>0, m>4 or m<-4x*x+mx+4 0

Let the inequality be x finch socks + mx + 4 = 0

That is, the equation has two roots, which are x1=(-b+ b -4ac) 2a=(-m+ m -16) 2, x2=(-b- b -4ac) 2a=(-m- m -16) 2, m -16 0, and m 4 or m -4 is obtained;

x1-x2=(-m+√m²-16)/2-(-m-√m²-16)/2=√m²-16≥0

x1 x2 is excitation x (1,2).

x1>2,x2<1

m+ m -16) 2> Roll 2

m²-16>(4+m)≥0

i.e. m -16> (4+m).

4+m)≥0

m -4, m <-4 (rounded).

m-√m²-16)/2<1

m+ m -16>-2, m -16>-2-m 0 i.e. m -16>-2-m

2-m≥0m<-5,m≤2

In summary, m>4

=b -4ac=m -16>0, m>4 or m<-4x*x+mx+4 0

Let the inequality be x + mx + 4 = 0

That is, the equation has two roots, which are x1=(-b+ b -4ac), Kai roll, 2a=(-m+ m -16) 2, x2=(-b- b -4ac), 2a=(-m- m -16) 2, m -16 0, and m 4 or m -4 is obtained;

x1-x2=(-m+ m -16) 2-(-m- handicap m -16) 2= m -16 0

x1≥x2x∈(1,2)

x1>2,x2<1

m+√m²-16)/2>2

m²-16>(4+m)≥0

i.e. m -16> (4+m).

4+m)≥0

m -4, m <-4 (rounded).

m-√m²-16)/2<1

m+ m -16>-2, m -16>-2-m 0 i.e. m -16>-2-m

2-m≥0m<-5,m≤2

In summary, m>4

x is not 0 and is greater than 0

then m1 then m<(x+2 x)min

Let y=x+2 x

y'=1-2/x^2

Lingy'=0 then starvation x=- root number 2 (rounded) or x = root imitation limb knowledge sign to eliminate 2ymin = 2 root number 2

m<2 root number 2

Let f(x)=mx -2x-1<0

The parabolic opening must be downward, i.e. m<0

Axis of symmetry x = 1 m

Then f(x) max = f(1 m) = -1 m-1<0 to get m<-1

In summary: m<-1

When m=0, it is not true.

When m is greater than 0, let the left side of the inequality be the function f(x), that is, there is an opening of f(x) upward, so it cannot be true.

When m is less than 0, the f(x) opening is downward.

It is sufficient to find the discriminant formula of the equation f(x)=0 less than 0.

That is, 2*2+4*1*m<0

i.e. m<-1

mx^2-2x-1<0

mx^2<2x+1

Because x belongs to r

Therefore, x 2 > = 0, when x = 0, the inequality is constant, m belongs to r, and when x is not equal to 0, the inequality is sorted into m< (2x+1) x 2 to find the minimum value of (2x+1) x 2.

Constant formation then m<0 opening downward vertices are (2 m,(-m+1) m) (m+1) m<0 m<0 1-m>0 m<1 composite m<0

According to the title, there are:

f(x)=x^2+mx+4

Then: f(1)<0

f(2)<0

So: 5+m<0

2m+8<0

i.e. m (-5].

This question should be parameter-determinated, ie.

m<-4-x2 x is derived from -4-x2 x, and -x2+4 x2 is obtained to find the minimum value of -x2+4 x2, and since this function is increasing on (1,2), the minimum value is taken when x=1.

Hence m<5