For the even function f (x) defined on [- 2,2], when x > = 0, f (x) decreases, then the inequality f (1-x)

For the even function f (x) defined on [- 2,2], when x > = 0, f (x) decreases, then the inequality f (1-x)

The even function f (x) defined on is monotonically decreasing on the interval,
Since f (x) is an even function, then f (- x) = f (x), the function image is symmetric about the Y axis,
So f (x) increases monotonically in the interval
f(1-m)

It is known that the even function f (x) defined on R decreases monotonically on [0, + ∞], and f (2x-1) > = f (x) is an inequality 2m + 1 / m

It is known that f (x) [0, + ∞] is monotonically increasing
And f (x) is even, so f (x) [- ∞, 0] is monotonically decreasing
In conclusion:
1. When 2x-1 ≥ 0, i.e. x ≥ 1 / 2, if f (1) < f (2x-1), then,
Because 1 > 0, 2x-1 ≥ 0
So if 2x-1 > 1, 2x > 1
X > 1
2. When 2x-1

If the odd function f (x) monotonically decreases on (- ∞, 0], then the solution set of the inequality f (lgx) + F (1) > 0 is______ ．

∵ the odd function f (x) monotonically decreases on (- ∞, 0]; f (x) monotonically decreases on [0, + ∞), that is, f (x) monotonically decreases on R. from F (lgx) + F (1) > 0, f (lgx) > - f (1) = f (- 1), ∵ lgx < - 1, the solution is 0 < x < 110, that is, the solution set of the inequality is (0110), so the answer is: (0110)

It is known that the square of 2008 times (x + y) and 2007 | 1 / 2 Y-1 | are opposite numbers, and the power of 2008 of x-2007 of Y is calculated

According to the meaning of the title, 2008 * (x + y) square = - 2007|1 / 2y-1|
The square is greater than or equal to 0, the absolute value is greater than or equal to 0, so the two items in the question can not be less than 0, can only be 0
So we know that x + y = 0,
1/2y-1=0
By solving the equation, y = 2, x = - 2
The problem is to find the 2008 times of (- 2) - (2) 2007 times = 2 2007 times (2-1) = 2 2007 times

If a and B are nonzero vectors, then a + b > A-B,

If the directions of vectors a and B are opposite, then | a + B | A-B |,

In △ ABC, O is the angle ABC, the intersection of the bisectors of angle ACB, OD ⊥ BC intersects D, the perimeter of △ ABC is 20, OD = 5, and the area of triangle ABC is calculated

Area = (5 * AB + 5 * BC + 5 * AC) / 2 = 5 * 20 / 2 = 50 do you understand

Y = - x2-3x + 1 (- 2 ≤ x ≤ 1) find the maximum and minimum of function

A:
y=-x^2-3x+1
The opening is downward and the axis of symmetry x = - 3 / 2
When x = - 3 / 2, the maximum value is - 9 / 4 + 9 / 2 + 1 = 13 / 4
When x = 1, the minimum value is - 1-3 + 1 = - 3

Mathematical proof of geometric problems
In the parallelogram ABCD, point E is on ad, connecting be, DF is parallel to be, BC is on F, AF and be are on M, CE and DF are on n
Verification: the quadrilateral mfne is a parallelogram

Because parallelogram
So ad parallels BC
Because be is parallel to DF
So the quadrilateral BEDF is a parallelogram
So BF = ed
So AE = CF
Angle DAF = angle BCE angle AEB = angle CFD
So the triangle ame is equal to the triangle FCN
So me = NF
So the quadrilateral mfne is a parallelogram

In a triangle, the bisectors of angle ABC and angle ACB intersect at point O. try to explain that the angle BOC = 90 ° + half angle A

∠BOC
= 180°-(∠OBC+∠OCB)
= 180°-(1/2)(∠ABC+∠ACB)
= 180°-(1/2)(180°-∠A)
= 90°+(1/2)∠A

x: 16 = 15: y = Z: 80 = 125% to solve the unknown

x=16x1.25=20
y=15/1.25=12
z=80x1.25=100