When x ∈ R, the univariate quadratic inequality x2-kx + 1 ＞ 0 holds, then the value range of K is______ ．

When x ∈ R, the univariate quadratic inequality x2-kx + 1 ＞ 0 holds, then the value range of K is______ ．

When ∵ x ∈ R, the univariate quadratic inequality x2-kx + 1 ＞ 0 holds, k2-4 ＜ 0, ∵ 2 ＜ K ＜ 2, so the answer is: - 2 ＜ K ＜ 2

For X ∈ [0,1], the inequality 1 / √ 1 + X ≤ 1-kx holds, then the value range of K is constant

Let T ^ 2 = x + 1  t ∈ [1, √ 2] x = T ^ 2-1, so the original inequality is changed to 1 / T ≤ 1-k (T ^ 2-1) (1) when t = 1, 1 ≤ 1, K takes any value (2) when t ≠ 1, the original inequality is changed to (T ^ 2-1) k ≤ 1-1 / T, that is, K ≤ 1 / T-1 / (T + 1), that is to find the right min, and the right min is 1 - √ 2 / 2. In conclusion, K ∈ [- ∞, 1 - √ 2 / 2]

If the inequality | x + 1 | ≥ KX holds for X ∈ R, then the value range of K is______ ．

∵ the inequality | x + 1 | ≥ KX holds, and the image of ∵ y = | x + 1 | cannot be below the image of y = KX. As shown in the figure, draw the images of two functions y = | x + 1 | and y = KX. According to the relationship between two straight lines, it is obtained that the image of y = KX can only be coincident with the X axis and parallel to y = x, ∵ 0 ≤ K ≤ 1, so the answer is: [0, 1]

For any a ∈ [1,3], the inequality ax ^ 2 + (A-2) X-2 > 0 holds and the range of X is obtained

ax^2+(a-2)x-2>0
(ax-2)(x+1)>0
When AX-2 > 0, x + 1 > 0,
x>2/a∩x>-1
Because a ∈ [1,3]
So x > 2
When AX-2

If a ∈ [- 2,2] inequality x ^ 2 + ax-3 ≥ a holds, find the range of X

X ^ 2-3 ≥ a (1-x), if x is less than 1, i.e. 1-x is greater than 0, then (x ^ 2-3) / (1-x) ≥ a, Let f (x) = (x ^ 2-3) / (1-x), find the derivative, then f (x) monotonically increases in [- 1,3], monotonically decreases in [3, positive infinity] or (negative infinity, - 1]. If x is less than 1, we can know that f (x) monotonically increases in [- 1,1] and monotonically decreases in (negative infinity, - 1]. If f (x) ≥ 2, then f (x) = 2, The solution is x = (radical 6) - 1 or - 1-radical 6, that is, X is less than or equal to - 1-radical 6. Similarly, if x is greater than 1, X is greater than 1 and less than or equal to 1 + radical 2. When x = 1, the inequality is 1 + A-3 ≥ a, which obviously does not hold. To sum up, X ∈ (negative infinity, - 1-radical 6] and (1,1 + radical 2]

Given that x = - 4 is a value in the solution set of inequality ax > 9, then the value range of a is______ ．

∵ x = - 4 is a solution of the inequality ax > 9, a < 0, the solution set of the inequality ax > 9 is: x < 9a, then - 4 < 9a, the solution is: a < - 94. So the answer is: a < - 94

A and B are all positive numbers, a + B = 1 prove the inequality that a multiplied by the square of X + B multiplied by the square of Y is greater than or equal to (AX + by)

In order to prove ax ^ 2 + by ^ 2 > = (AX + by) ^ 2, that is to say, ax ^ 2 + by ^ 2 - (AX + by) ^ 2 > = 0ax ^ 2 + by ^ 2 - (AX + by) ^ 2 = ax ^ 2 + by ^ 2 - (AX) ^ 2-2abxy - (by) ^ 2 = a (1-A) x ^ 2 + B (1-B) y ^ 2-2abxy, according to the known a + B = 1 = ABX ^ 2 + aby ^ 2-2abxy = AB (x ^ 2-2xy + y ^ 2), use the complete square

Ax + by ≥ (AX + by)

ax^2+by^2≥(ax+by)^2
∵ax^2+by^2-(ax+by)^2
∴=ax^2+by^2-(ax^2+2axby+by^2)
=2axby≥0
∴ax^2+by^2≥(ax+by)^2
^2 is the square

Let a > 1, prove that the inequality (1 + x) ^ a > 1 + ax holds when x > 1

（1+x)^a-(1+ax)>X+aX-1-aX=X-1,
When x > 1, X-1 > 0,
So the original formula is proved
Just expand the first two terms with binomial theorem

We know the solution set of inequality ax + B0 about X

The same solution of ax1
a