Let a, B, x, t satisfy ax + by = 7, ay BX = 10, Then the value of (a square + b square) (x square + y Square) is

Let a, B, x, t satisfy ax + by = 7, ay BX = 10, Then the value of (a square + b square) (x square + y Square) is

The original formula = A & sup2; X & sup2; + A & sup2; & sup2; Y & sup2; + B & sup2; X & sup2; + B & sup2; Y & sup2;
=(a²x²+2abxyb²y²)+(a²y²-2abxy+b²x²)
=(ax+by)²+(ay-bx)²
=7²+10²
=149
If a, B, x, y satisfy ax + by = 3, ay-bx = 5, then the value of (A & # 178; + B & # 178;) (X & # 178; + Y & # 178;) is
ax+by=3,ay-bx=5
Square addition
a²x²+b²y²+a²y²+b²x²=34
That is (A & # 178; + B & # 178;) (X & # 178; + Y & # 178;) = 34
Given that real numbers a, B, X and y satisfy A2 + B2 = 1, X2 + y2 = 3, then the maximum value of AX + by is___ .
Because A2 + B2 = 1, X2 + y2 = 3, from Cauchy inequality (A2 + B2) (x2 + Y2) ≥ (AX + by) 2, we get 3 ≥ (AX + by) 2. If and only if ay = BX, we take the equal sign, so the maximum value of ax + by is 3. So the answer is: 3
Given the real number x, y, a, B, satisfying x ^ 2 + y ^ 2 = 1, a ^ 2 + B ^ 2 = 3, then the maximum value of AX + by is
The addition of two inequalities may enlarge the scope, which should be avoided in the process of solving the problem
(x ^ 2 + y ^ 2) (a ^ 2 + B ^ 2) = 3 = a ^ 2x ^ 2 + B ^ 2Y ^ 2 + A ^ 2Y ^ 2 + B ^ 2x ^ 2 ≥ a ^ 2x ^ 2 + B ^ 2Y ^ 2 + 2abxy = (AX + by) ^ 2, that is, 3 ≥ (AX + by) ^ 2, root 3 ≥ ax + by
Let the domain of definition of function f (x) be d. if for any x ∈ D, there exists y ∈ d such that [f (x) + F (y)] / 2 = C (C is a constant), then the mean value of function f (x) on D is called C. the following four functions are given: 1, y = x ^ 3,2, y = (1 / 2) ^ X; 3, y = LNX; 4, y = 2sinx + 1, then the number of functions satisfying the mean value of 1 on its domain is___________
The definition field is inconsistent with the value field and does not meet the requirement of "on its definition field...". It is wrong to say that there exists y
1,3,4
2 is wrong, y = (1 / 2) ^ x > 0
When x is a small negative number, such as - 5, there is no corresponding y
Only y = 2sinx + 1 is a function.
When the domain D is a point on the X axis, the condition is satisfied.
The definition of its range is not consistent with that of other functions.
(2) it doesn't exist.
[summary]
① To prove compliance, it must be proved to be consistent with all the circumstances;
② On the contrary, it can be proved that there is only one case of non conformity.
The mission passed by
If the image of a quadratic function passes through (1,4) and (5,0), and the symmetry axis X = 2, try to find the expression of the quadratic function
Axis of symmetry x = 2
y=a(x-2)²+k
It's two o'clock
be
4=a+k
0=9a+k
subtract
8a=-4
a=-1/2
k=-9a=9/2
So y = - (X-2) & sup2 / 2 + 9 / 2
That is y = (- X & sup2; + 4x + 5) / 2
Define the function y = f (x), X ∈ D, if there is a constant C, for any x1 ∈ D, there is a unique x2 ∈ D, such that f (x1) + F (x2) 2 = C, then the mean value of function f (x) on D is C. given that f (x) = lgx, X ∈ [10100], then the mean value of function f (x) = lgx on X ∈ [10100] is C___ .
According to the definition, the function y = f (x), X ∈ D, if there is a constant C, for any x1 ∈ D, there is a unique x2 ∈ D, such that f (x1) + F (x2) 2 = C, then the mean value of function f (x) on D is called C. let x1 · x2 = 10 × 100 = 1000, when x1 ∈ [10100], select x2 = 1000x1 ∈ [10100], then C = 12L
It is known that the symmetry axis of quadratic function is a straight line x = - 2, and the image passes through points (1,4), (5,0), and the expression of quadratic function is obtained
Let the quadratic function be y = a (x + 2) &# 178; + K, then
9a+k=4
49a+k=0
a=-0.1
k=4.9
The quadratic function expression is y = - 0.1 (x + 2) &# 178; + 4.9 = - 0.1X & # 178; - 0.4x + 4.5
If the axis of symmetry is x = - 2, the function equation can be set as y = a (x + 2) &# 178; + C
The function expression can be obtained by bringing in points (1,4), (5,0)
y=-0.1（x+2）²+4.9
High school mathematics: F (x) = (2a-1 / a) - (1 / a square x), constant a > 0
F (x) = (- 1 / A & sup2;) / x + (2a + 1) / A, (a > 0) is an increasing function in the interval [M, n]. F (x) min = f (m) = (- 1 / A & sup2;) / m + (2a + 1) / a = M
Given that the image of quadratic function y = f (x) passes through point (0,3), the symmetry axis of the image is x = 2, and the difference between the two zeros of y = f (x) is 2, the analytic expression of y = f (x) is obtained
Let f (x) = AX2 + BX + C & nbsp; (a ≠ 0), ∵ f (x) pass through point (0,3), ∵ C = 3; and the symmetry axis of F (x) is x = 2, ∵ B2A = 2, that is, B = - 4A, ∵ f (x) = ax2-4ax + 3 (a ≠ 0); let the two real roots of the equation ax2-4ax + 3 = 0 (a ≠ 0) be & nbsp; If x 1, x 2, and x 1 ＞ x 2, then according to the meaning of the title: x 1 + x 2 = 4, x 1 x 2 = 3A, x 1 − x 2 = 2, x 1 = 3, x 2 = 1, x 3 a = x 1 x 2 = 3; a = 1, B = - 4; y = f (x), the analytic formula is f (x) = x 2-4x + 3