A math problem in the first year of high school is very simple

Inequality 1 [2 (x 2+x)]>1 2) (2x 2-mx+m+4) holds constantly.

It can be seen that x 2+x> 2x 2-mx+m+4 is always established.

x 2-(m+1)x+m+4<0 is established.

For the quadratic function y=ax 2+bx+c, its vertex coordinates are (-b 2a, (4ac-b 2) 4a), i.e.

h=-b/2a=(x1+x2)/2

k=(4ac-b^2)/4a

So (m+1) (m+4) <0 is constant.

4(m+4)-(m+1) 2] 4 <0 Constant.

Obviously, m= =4 solution (m+1) (m+4) <0 gives -4, so -4

1/2)^(x^2-x) >1/2)^(2x^2-mx+m+4)x^2-x > 2x^2-mx+m+4

3x^2+(1-m)x+m+4 <0

The curve opening is upward, if the above equation is greater than 0, then let the discriminant formula be less than 0 and the vertex is less than 0, then you can find it. And now that the opening is upward, there is no eligible m, of course, it is possible that I miscalculated.

Vertex coordinates formula.

1 2) (x 2-x)> (1 2) (2x 2-mx+4) gives -x 2-x<2x 2-mx+4

i.e. 3x 2-(m-1)x+m+4>0

Vertex y=(4ac-b2) 4a>0

i.e. 12(+4)-(m-1) 2>0

m^2-14m-47>0

The solution. m>7+4 root 6

or m<7-4 root 6

t=-ab/|b|^2

The included angle is 90 degrees.

t=-ab/(|b|^2)

b|That's what mold means.

There are many ways to do this.

For example, let y=a+bt

y^2=a^2+(bt)^2+2abt

This is a quadratic equation with t as a variable.

And the value of y must be greater than 0, so t=-ab |b|2 That's fine, you can also use the way you draw, etc.

When doing physics, the problem of crossing the river is the same, take a closer look at it, and you can know that u*b=0 by substituting the t value, that is, the angle between a+bt and b is 90

Let k=x+y, y=k-x be substituted into:

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

2x^2-2kx+k^2-1=0

If the equation has a solution, and if the discriminant equation 0 is, then there is:

4k^2-8(k^2-1)>=0

4k^2+8>=0

k^2<=2

Root number 2 k root number 2

i.e. the minimum root number 2

Passing the point C as CE AB perpendicular foot is E, because AB is the diameter of the circle, so ABC is RT, which is obtained by the projective theorem: BC 2=Be*AB, that is, X 2=BE*2R, BE=X 2 (2R), so the top bottom CD=AB-2*BE=2R-2*X 2 (2R)=2R-X 2 R, so Y=2R+2R-X 2 R+2X, that is, Y=4R-X 2 R+2X

0< the arithmetic square root of x<2 *r< p>