If the absolute value of the square of (y + Z + 2x-y) is not equal to the square of (y + Z + 2x-y), then the absolute value of (y + y) is not equal to the square of (y + Z-Y), If the absolute value of 2x-3y-8z + X + y + Z = 0, and Z is not equal to 0, then what is the value of XZ square + YZ square * (xsquare-y).

If the absolute value of the square of (y + Z + 2x-y) is not equal to the square of (y + Z + 2x-y), then the absolute value of (y + y) is not equal to the square of (y + Z-Y), If the absolute value of 2x-3y-8z + X + y + Z = 0, and Z is not equal to 0, then what is the value of XZ square + YZ square * (xsquare-y).

|2x-3y-8z|+|x+y+z|=0
Note: 2x-3y-8z = 0 and X + y + Z = 0,
From this we can get the following conclusion;
x=z,y=-2z;
(xz^2+yz^2)(x^2-y)/x
=z^2(x+y)(z^2+2z)/z
=-z^2(z^2+2z).
Let x, y ∈ R, a ＞ 1, B ＞ 1, if AX = by = 3 (x, y are exponents), a + B = 2 √ 3, then the maximum value of (1 / x) + (1 / y) is
I want to know the process
Ax = by = 3 (x, y are exponentials), then x = loga 3, y = logb 3 is exponentialized, right? 1 / x + 1 / y = 1 / loga 3 + 1 / logb 3 = 1 / (LG 3 / LG a) + 1 / (LG B / LG 3) here, we use the formula (logm n = LG n / LG m) = LG A / LG 3 + LG B / LG 3 = LG a * B / LG 3, because a * B ≤ ((a + b) / 2) square = 3
X = log (a, 3) y = log (B, 3)
Then (1 / x) + (1 / y) = log (3, a) + log (3, b) = log (3, ab)
And because a + b > = 2 √ AB (a > 1, b > 1) (basic inequality 2)
So ab
It is known that: A ^ 2 + B ^ 2 = 1, x ^ 2 + y ^ 2 = 1, then the maximum value of AX + by is?
Ax
Let a = (a, b) B = (x, y)
|Vector a | = 1
|Vector B | = 1
Vector a * vector b = | vector a | * | vector B | * cosa
When x > 0, if the value of function f (x) = (3a-2) x is always greater than 1, then the value range of real number a is ()
A. （23，1）B. （-∞，1）C. （1，+∞）D. （0，23）
When x ＞ 0, (3a-2) x ＞ 1 = (3a-2) 0; ｜ the exponential function should be an increasing function; ｜ 3a-2 ＞ 1; ｜ a ＞ 1, the range of real number a is: (1, + ∞). So the answer is: (1, + ∞)
The equation of symmetry axis of quadratic function y = x ^ 2-4x + 5 image is
y＝x^2－4x＋5
=(x-2)^2+1
The equation of axis of symmetry is: x = 2
pass by
X = 2. The equation of axis of symmetry is x = - B / 2A. Y = ax ^ 2 + BX + C can be formulated
The function f (x) is a power function, and the image passes through the point (2,8). The function y = f (x) defined on the real number R is an odd function. When x > 0, f (x) = f (x) + 1, find the expression of F (x) on R, and draw the image
Let y = x α, (x > 0); substituting (2, 8) into α = 3, when x > 0, f (x) = f (x) + 1 = X3 + 1, when x < 0, - x > 0, f (- x) = (- x) 3 + 1 = - X3 + 1, ∵ y = f (x) is odd function, ∵ f (- x) = - f (x) ∵ f (x) = x3-1, ∵ y = f (x) is odd function defined on real number r, ∵
The symmetry axis equation of the image of quadratic function y = x square + 2x + 2
For the quadratic function y = ax ^ 2 + BX + C, the axis of symmetry is x = - B / 2A
So the answer is: x = - 1
x=-1
x=-1
If the function f (x) = (2x + 1) (x + a) x is odd, then the value of real number a is___ .
F (- x) = (- 2x + 1) (- x + a) - x = 2x2 - (2a + 1) x + A-X = - 2x2 + (2a + 1) x + ax; ∧ 2x2 - (2a + 1) x + a = 2x2 + (2a + 1) x + A; ∧ 2A + 1) = 2A + 1, ∧ a = - 12
Given that the image of the power function y = x ^ ((m ^ 2) - 2m-3) (m ∈ n +) does not intersect the coordinate axis and is symmetric about the Y axis, then M =?
The answer is 1 or 3, but if it is 3, if the power function is y = x, it will be called at the origin?
I don't understand that,
The answer is 1,
3 is not consistent
With respect to Y-axis symmetry, it's an even function, so the exponent is even
And x-axis, Y-axis have no intersection
Then its image is similar to the inverse scale function in a quadrant
That is, the index is less than 0
2m-2 M-3
0 has no 0 power
Let f (x) = x ^ 2-2x, and let a satisfy | x-a|
|f(x)-f(a)|=|x^2-2x-a^2+2a|
=|(x-a)(x+a)-2(x-a)|
=|x-a|*|(x+a)-2|