It is known that the lengths of three sides of a triangle are 5cm, 8cm and 7cm. If a triangle can be formed by cutting them off xcm, then the value range of X is?

It is known that the lengths of three sides of a triangle are 5cm, 8cm and 7cm. If a triangle can be formed by cutting them off xcm, then the value range of X is?

5-x+7-x>8-x
x

Solution equation: 4 (X-2) ^ 2 = 9 9x ^ 2-6x + 1 = 0

4(x-2)^2=9
(x-2)^2=9/4
x-2=±3/2
x=2+3/2=3.5
x=2-3/2=0.5
9x^2-6x+1=0
(3x-1)^2=0
3x-1=0
x=1/3

In p-abc, PA, Pb and PC form an angle of 60 degrees, PA = a, Pb = B, PC = C. the volume of a triangular pyramid is calculated

Let a > b, a > C, then we can construct a regular tetrahedron p-amn, where B is on PM and C is on PN; we can first calculate the volume of regular tetrahedron, and then calculate the volume of PABC according to V (PAMN) / V (PABC) = PA / PA * PM / Pb * pN / PC

8x-1 / 3 = 5 / 9 (solving equation)
5 minutes plus reward

8x-1 / 3 = 5 / 9
8x=5/9+1/3
8x=5/9+3/9
8x=8/9
x=1/9
X = 1 / 9

If the real number x + y = 1, then the value range of 1 / x + X / y is calculated. The answer is greater than or equal to 3 or less than or equal to - 1

X + y = 1, then:
M=1/x＋x/y
=(x＋y)/x＋x/y
=1＋[(y/x)＋(x/y)]
1. If x and y have the same sign, then x / Y and Y / X are both positive, then m ≥ 1 + 2 = 3
2. If x and y are different signs, then x / Y and Y / X are negative numbers, then
If (x / y) - (Y / x) ≥ 2, then m ≤ - 1
Thus, there are: m ≥ 3 or m ≤ - 1

Given that a and B are opposite numbers, C and D are reciprocal, and the square of M = 49, then M = CD, what is a + B?

Because AB is opposite to each other, CD is reciprocal to each other
So a + B = 0, CD = 1
Because M = a + B / CD, the square of M = 49
So m = 7
There is something wrong with this question!

Given that a and β are the two real roots of the equation x ^ 2-2ax + 6x + a = 0, find the minimum value of (A-1) ^ 2 + (β - 1) ^ 2

α. β is the two real roots of the equation x & # 178; - 2aX + 6x + a = 0, that is, α, β is the two real roots of the equation x & # 178; - (2a-6) x + a = 0, then α + β = 2a-6, α β = a (α - 1) &# 178; + (β - 1) &# 178; = α & # 178; - 2 α + 1 + β & # 178; - 2 β + 1 = α & # 178; + β & # 178; - 2 α - 2 β + 2 = (α + β) &# 178; - 2 α β - 2

If the function f (x) = x3-3x + A has three different zeros, then the value range of real number a is ()
A. （-2，2）B. [-2，2]C. （-∞，-1）D. （1，+∞）

The solution ∵ f ′ (x) = 3x2-3 = 3 (x + 1) (x-1), when x ＜ - 1, f ′ (x) ＞ 0; when - 1 ＜ x ＜ 1, f ′ (x) ＜ 0; when x ＞ 1, f ′ (x) ＞ 0; when x = - 1, f (x) has a maximum. When x = 1, f (x) has a minimum. In order to make f (x) have three different zeros, we only need f (− 1) ＞ 0f (1) ＜ 0, the solution is - 2 ＜ a ＜ 2

If the solution set of the inequality system x-a ＞ 2 (1) b-2x ＞ 0 (2) is - 1 ＜ x ＜ 1, then the 2011 power of (a + b) =?
If the solution set of the inequality system x-a ＞ 2 ① is - 1 ＜ x ＜ 1, then the 2011 power of (a + b) =? B-2x ＞ 0 ②

-1

Is the general form of quadratic function y = ax & # 178; + BX + C the y value of vertex coordinates?
I have done a problem, y and C are not equal. For example, if we know the parabola y = x & # 178; - 4x + 3, we can calculate y = (4ac-b & # 178;) / 4A, y = 11 / 4 by formula method. Is it true that only the vertex K is equal to y

C is not the y value of the vertex coordinate, but the y value of the intersection with the Y axis