If a, B, X and y satisfy ax + by = 4 and ay BX = 5, then (a ^ 2 + B ^ 2) + (x ^ 2 + y ^ 2)=

If a, B, X and y satisfy ax + by = 4 and ay BX = 5, then (a ^ 2 + B ^ 2) + (x ^ 2 + y ^ 2)=

Is it like this
(a^2+b^2)(x^2+y^2)
=a^2*x^2+b^2*y^2+a^2*y^2+b^2*x^2
=(ax)^2+(by)^2+2abxy+(ay)^2+(ba)^2-2abxy
=(ax+by)^2+(ay-bx)^2
=4^2+5^2
=41
If a, B, X and y are all positive real numbers and X + y = 1, we prove that ab ≤ (AX + by) (ay + BX) ≤ (a + b) 24
This is the proof that: let (a (x + by) (ay + BX) - AB = a (a (x + by) (ay + BX) - AB = a (a (x + by) (ay + by) (ay + by) (ay + by) (a (x + y) 2-2xy-1] \\ (a + b) (ay + BX) (a (x + by) (ay + BX) - AB = a (a (x + by) (a (x + by) (a + B (x + b) (a + B (x + B) (a + b) (a + B (x + b) (a + B (x + by) (a + by) (a + by) (a) and (AX + by) (a) and (a (x + by) (a) and (x + by) (a) and (a (x + by) (also) (a) (a) (a) (x + by) (a) (x + by) (x + by) (a) (x + by) (x + by) (a) (x + by) (x + by) (a) (a 2 = (a + B2) 2 = (a + b) )24．∴ab≤（ax+by）（ay+bx）≤(a+b)24．
The real numbers a, B and X, y satisfy the condition a
（ax+by）-（ay+bx）
=by-ay-bx+ax
=（b-a）·y-（b-a）·x
=（b-a）·（y-x）
>0
So: ax + by > ay + BX
You can also ask me questions here:
If the image of quadratic function passes through the far point and (- 4,0), then the equation of symmetry axis of quadratic function image is?
x=(0+(-4))/2=-2
The axis of symmetry is:
x=-2.
The midpoint of (0,0) and (- 4,0) is (- 2,0), so the axis of symmetry x = - 2
Quadratic function y = ax ^ 2 + BX + C (a is not equal to 0)
After the origin and (- 4,0) C = 0, the equation is y = ax ^ 2 + B
And 16A + B = 0
16a=-b
-b/2a=8
So, the axis of symmetry equation is
X=8
That is to say, there are two intersections between the quadratic function and the x-axis, that is, there are two solutions, then the axis of symmetry is the midpoint of the two solutions.
That is: [0 + (- 4)] / 2 = - 2, the axis of symmetry is x = - 2.
Let f (x) = x ^ 2 - (3a-1) x + A ^ 2 increase in [1, + ∞), then the value range of the real number a
∵ the function f (x) = x ^ 2 - (3a-1) x + A ^ 2 is an increasing function in [1, + ∞),
The function f (x) = x ^ 2 - (3a-1) x + A ^ 2 is open upward, and the axis of symmetry is x = (3a-1) / 2,
Then x = (3a-1) / 2 ≤ 1, i.e. a ≤ 1, the value range of real number a is (- ∞, 1]
As long as the axis of symmetry of the function is greater than or equal to 1
If the image of the quadratic function passes through the origin and the point (- 4,0), then the equation of the symmetry axis of the image of the quadratic function is? And needs a complete solution,
Solving the equation of axis of symmetry
The image of a quadratic function has an obvious feature, that is, the abscissa of its axis of symmetry is half of the intersection of its two axes
If the function image passes through the origin and (- 4,0), then the abscissa of the axis of symmetry must be (0 + (- 4)) / 2 = - 2
So the axis of symmetry equation is x = - 2
That is, when x = 0 and x = - 4, y is 0
So the axis of symmetry is x = (- 4 + 0) / 2
That is x = - 2
If the function f (x) = - X & # 178; + (3a-1) x + 2a is a decreasing function in the interval (- ∞, 4), then the value range of real number a
A a≤-3 B a≥3 C a≤5 D a=-3
F (x) = - X & # 178; + (3a-1) x + 2A has a downward opening and an increasing function on the left side of the axis of symmetry
-Infinity must be on the left side of the axis of symmetry, and it can't be a decreasing function on the interval (- ∞, 4), so it's wrong!
Suppose f (x) = x & # 178; + (3a-1) x + 2a is a decreasing function in the interval (- ∞, 4)
Opening up
Axis of symmetry x = (3a-1) / (- 2) = (1-3a) / 2
The interval is to the left of the axis of symmetry
4≤(1-3a)/2
1-3a≥8
-3a≥9
a≤-3
The - B below him is wrong
Because x > 0, the axis of symmetry is on the right side of 4,
Then: (formula of axis of symmetry)
-b/2a≤4
That is: - (3a-1) / 2 ≤ 4
A ≤ - 3
The results are as follows:
-b/2a≤4
That is: (3a-1) / 2 ≤ 4
D: a ≤ 3 - follow up: - no such option
The axis of symmetry equation of quadratic function y = x2-2x + 1 is______ .
∵-b2a=-−22=1∴x=1．
Whether there is such a real number a that the function f (x) = x2 + (3a-2) x + A-1 has an intersection with the x-axis in the interval [- 1,3], and there is only one intersection. If there is, find out the range. If not, explain the reason
If the real number a satisfies the condition, it only needs f (- 1) · f (3) ≤ 0. F (- 1) · f (3) = (1-3a + 2 + A-1) · (9 + 9a-6 + A-1) = 4 (1-A) (5a + 1) ≤ 0. So a ≤ - 15 or a ≥ 1. Test: (1) when f (- 1) = 0, a = 1. So f (x) = x2 + X. Let f (x) = 0, that is, X2 + x = 0. Then x = 0 or x = - 1. There are two equations on [- 1, 3], so a ≠ 1. (2) When f (3) = 0, a = - 15, then f (x) = x2-135x-65. Let f (x) = 0, that is, x2-135x-65 = 0, the solution is x = - 25 or x = 3. There are two equations on [- 1, 3], so a ≠ - 15
The axis of symmetry equation of quadratic function y = x-2x + 3 is
y=x^2-2x-3=(x-1)^2-4
The axis of symmetry is x = 1
Symmetry axis calculation - (B / 2a) = - (- 2 / 2 × 1)
＝ 1
So the axis of symmetry x = 1
X = 1 is the child just learning here? There are formulas in the textbook