It is known that ax ^ 3 = by ^ 3 = CZ ^ 3 and 1 / x + 1 / y + 1 / z = 1. Prove that: (AX ^ 2 + by ^ 2 + CZ ^ 2) ^ 1 / 3 = a ^ 1 / 3 + B ^ 1 / 3 + C ^ 1 / 3

It is known that ax ^ 3 = by ^ 3 = CZ ^ 3 and 1 / x + 1 / y + 1 / z = 1. Prove that: (AX ^ 2 + by ^ 2 + CZ ^ 2) ^ 1 / 3 = a ^ 1 / 3 + B ^ 1 / 3 + C ^ 1 / 3

Let ax ^ 3 = by ^ 3 = CZ ^ 3 = s ^ 3,
∴(ax^2+by^2+cz^2)^1\3
=(s^3/x+s^3/y+s^3/z)^1/3
=[s^3(1/x+1/y+1/z)]^1/3
=s
∵a^1\3+b^1\3+c^1\3
=s/x+s/y+s/z
=s(1/x+1/y+1/z)
=s
∴(ax^2+by^2+cz^2)^1\3=a^1\3+b^1\3+c^1\3.
Let x = by + CZ, y = CZ + ax, z = ax + by, find the value of (A / A + 1) + (B / B + 1) + (C / C + 1)
x=by+cz
ax=y-cz
=>(a+1)x=(b+1)y
The same is true
(c+1)z=(b+1)y
(a+1)x=(c+1)z
The original formula = ax / (B + 1) y + by / (B + 1) y + CZ / (B + 1) y
=(ax+by+cz)/(b+1)y
=(z+cz)/(b+1)y
=1
Let the solution set of quadratic inequality ax ^ 2 + BX + 1 > 0 be {X - 1 ＜ x ＜ 3 / 1}, then a =?, B =?
A0
Let ax ^ 2 + BX + 1 = 0
x1=[-b-(b^2-4a)^(1/2)]/2a
x2=[-b+(b^2-4a)^(1/2)]/2a
x1=-1
x2=3/1
a=-1/3
b=2/3
Because the solution set of inequality ax ^ 2 + BX + 1 > 0 is {X - 1 < x < 3 / 1}, the solution of equation AX ^ 2 + BX + 1 = 0 is - 1,3
-b/a = -1 + 3 = 2
1/a = -3
The solution is as follows
a = -1/3
b= 2/3
a=-1/3 b=2/3
First, by - 1
The range of y = (3x + 2) / (X-2)
Domain x is not equal to 2
y=(3x+2)/(x-2)=(3x-6+8)/(x-2)=(3x-6)/(x-2)+8/(x-2)=3+8/(x-2)
Since 8 / (X-2) is not equal to 0, y is not equal to 3
y=(3x+2)/(x-2)=3+8/(x-2)
If the definition field is x and not equal to 2, the value field is not equal to 3
y=(3x+2)/(x-2)=
xy-2y=3x+2=
x(y-3)=2+2y=
X = 2 + 2Y / Y-3 because the denominator cannot be 0
So y can't be equal to 3
Take the formula apart and make it y = 3 + 8 / (X-2)
If there is no definition field of X, the value field of this formula is that y is not equal to 3
If the solution set of inequality ax ^ 2 + bx-2 > 0 is (1,2), what is the value of a ^ 2 + B ^ 2
Because the solution set of ax & # 178; + bx-2 > 0 is (1,2), that is, 1
A
Y = 3x ^ 2-x + 2. X ∈ [1,3] for range
Y = 3 (x-1 / 6) ∧ 2-23 / 12, about x = 1 / 6 symmetry, 1 / 6 is not in [1,3], so the range is [4,26]. Directly take 1,3 into the calculation
The quadratic function y = ax ^ 2 + BX + C (a) is known
The two zeros are - 5 / 2 and 1 / 2, respectively
∵ quadratic function y = ax ^ 2 + BX + C (a)
The quadratic function y = ax ^ 2 + BX + C (a) is known
Find the range of y = (x ^ 2 + 3x + 3) / (x + 1)
y=（x^2+3x+3）/（x+1）
x+1≠0
y(x+1)=x²+3x+3
x²+(3-y)x+(3-y)=0
The above quadratic equation of one variable with respect to X has real solutions
So there is △ > = 0
△=(3-y)²-4(3-y)>=0
(3-y)(-1-y)>=0
-1
It is known that the inequality (AX-5) (x ^ 2-A) A9 ∵ 5 about X does not belong to M. substituting x = 5, the inequality does not hold ∵ (5a-5) / (25-A) ≥ 0 or A-25 = 0 ∵ (A-1) / (A-25) ≤ 0 or A-25 = 0 = = > 1 ≤ a ≤ 25 ∵ ① the value range of the intersection number a is 1 ≤ a
sure
This is the denominator 0,
If we really find out x = 5, we can't go camping
So 5 doesn't belong to M
y=(4/5)(x^2-3x-1) 3
y=(4/5)(x^2-3x-1)
Because the axis of symmetry is x = 3 / 2
-4/5