-
Anonymous users2024-02-07Define the operation: x y=x(1-y).
This means substituting the pre-definition expression into the x position in x(1-y) and the corresponding expression in the y position in x(1-y), so that (x-a) (x+a)=(x-a)*[1-(x+a)]=(x-a)*(x-a+1).
This problem is: (x-a)*(x-a+1)<1 is true for any real number x, find the range of a.
x^2+ax-ax+a^2+x-a<1
Equivalent to: x 2-x-a 2+a+1>0
Equivalent to: (x-1 2) 2-a 2+a+3 4>0
When x takes 1 2, (x-1 2) 2-a 2+a+3 4 is the minimum value, that is, as long as x=1 2 satisfies (x-1 2) 2-a 2+a+3 4>0, then this inequality can be true regardless of the real value of x.
Therefore, (1 2-1 2) 2-a 2+a+3 4>0 is a 2-a-3 4<0
4a^2-4a-3<0
2a-3)*(2a+1)<0
1/2
-
Anonymous users2024-02-06x-a)(1-x-a)<1
Sorted out x-x -a+a -1<0
r to find the maximum value of x-x squared, and then calculate it yourself.
-
Anonymous users2024-02-05x-a）⊙(x+a)<1
x-a)(1-x-a)＜1
x²+x-a+a²-1＜0
x²-x+a-a²+1＞0
Let f(x)=x -x+a-a +1
This is a quadratic function.
The value of this function is always greater than 0
That is, the image is always above the x-axis.
It is also because the coefficient of the quadratic term is greater than 0
So the opening is upward.
Constant above the x-axis, i.e., no intersection with the x-axis.
The equation x -x + a-a + 1 = 0 has no solution.
1-4（a-a²+1）＜0
4a²-4a-3＜0
Decompose the factor. 2a+1)(2a-3)＜0
1/2
-
Anonymous users2024-02-04The left side of the inequality is equal to:
x-a）（1-x-a）
x-x²-ax-a+ax+a²
x²+x+a²-a
Formulation. -(x-1 2)square+a+a+1 4 i.e.g., find:-(x-1 2)square+a+1 4<1 after deformation: x-1 2)square-a+a+3 4>0 by the property of non-negative numbers:
To ensure that (x-1 2) squared -a +a+3 4>0 is true for any x, -a +a+3 4 must be made to be 0. i.e. -a +a+3 4>0
Solving the inequality yields a in the range of -1 2 to 2 3.
Remember to add points...
-
Anonymous users2024-02-03It is known that the coordinates of the three points are (-1,0)(3,-1)(1,2), AE vector=1 3AC vector, BF vector=1 3BC vector, and the EF vector and AB vector are collinear.
ac=(2,2)
bc=(-2,3)
ab=(4,-1)
ae=1/3ac=(2/3,2/3)
bf=1/3bc=(-2/3,1)
ef=ea+ab+bf=(-2/3,-2/3)+(4,-1)+(2/3,1)
8 3, -2 3) = 2 3 (4, -1) = 2 3 The ABEF vector is collinear with the AB vector.
From the known results, -1 is less than or equal to x+1 less than or equal to 1, and the solution is that 2 is less than or equal to x and less than or equal to 0 >>>More
0< 0, sin >0, sin >0coss >0 ,cos >0, cos >0sin( + sin( +sin sin must be proved. sin(θ+sinβ>sin(θ+sinαsinθcosα+sinαcosθ)sinβ>(sinθcosβ+sinβcosθ)sinα >>>More
Because x 2 + y 2 > = 2xy
y^2+z^2>=2yz >>>More
It seems that the first question is incomplete.
Let x2 > x1, and x1 and x2 both belong to [0, 2].
f(x2)-f(x1)=-2acos2x2+b+2acos2x1-b=2a(cos2x1-cos2x2) >>>More