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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
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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.
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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
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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
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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
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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>
a={x|0,-4}
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I think this question can be seen as yes.
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