For real numbers, a new operation is defined, X * y = ax + by + XY, where a and B are constants. It is known that 2 * 1 = 7, (- 3) * 3 = - 10, and the value of one third * 6 is obtained

For real numbers, a new operation is defined, X * y = ax + by + XY, where a and B are constants. It is known that 2 * 1 = 7, (- 3) * 3 = - 10, and the value of one third * 6 is obtained

x*y=ax+by+xy,
2*1=7,（-3）*3=-10
2a+b+2=7
2a+b=5
-3a+3b-9=-10
-3a+3b=-1
Solution
a=16/9
b=13/9
x*y=16x/9+13y/9+xy
1/3*6=16/27+26/3+2
=304/27
x*y=ax+by+xy
2*1=7，（-3）*3=-10
2a+b+2=7
-3a+3b-9=-10
Solution
a=16/9
b=13/9
(1/3)*6=a/3+6b+2=304/27
From X * y = ax + by + XY
2*1=2a+b+2=7
（-3）*3= -3a+3b-9= -10
To solve this system of linear equations, a = 16 / 9, i.e. 16 / 9, B = 13 / 9, i.e. 13 / 9
Then the values of a and B are substituted into the original formula
One third * 6 = (1 / 3) * a + 6B + (1 / 3) * 6 + 2 = 304 / 25
Because x * y = ax + by + XY, so 2 * 1 = 2A + B + 2 = 7, so 2A + B = 5. Similarly, (- 3) * 3 = - 3A + 3b-9 = - 10, so - 3A + 3B = - 1, the simultaneous equations get a = 16 / 9, B = 13 / 9, X * y = 16x + 13y + XY / 9, one third * 6 = 358 / 27, I don't know if it's right, but the method is right. Maybe the number is wrong, hope you can adopt it, thank you
Twenty-six
___ twenty-six-thirds
Three
Let a, B, x, y ∈ R and satisfy A2 + B2 = m, X2 + y2 = n, the maximum value of AX + by is ⊙___ .
From Cauchy inequality, we can know that (A2 + B2) (x2 + Y2) ≥ (AX + by) 2, that is, 1 ≥ (AX + by) 2, ax + by ≤ Mn, so the answer is: Mn
Given ax + by = 3, ay BX = 5, then the value of (A & # 178; + B & # 178;) (X & # 178; + Y & # 178;) is
(ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)=34
The symmetry axis of function f (a + x) = f (b-X) and its derivation
F (x + a) denotes that the function f (x) is shifted a unit to the left, f (b-X) denotes that the function f (x) is shifted B units to the left after flipping about the y-axis, and f (x + a) = f (b-X), that is, the function f (x) is shifted a unit to the left after flipping about the Y-axis and then shifted B units to the left, so the axis of symmetry is x = [(a + x) + (b-X)] / 2 = (a + b) / 2
Let real numbers a and B make equation X4 + AX3 + bx2 + ax + 1 = 0, and find the minimum value of A2 + B2
Let t = x + 1 x, then T2 + at + B-2 = 0, and t ≥ 2. Let g (T) = T2 + at + at + B-2, (124 ≥ 2). Let g (T) g (t (T) = T2 + 2 + 2 + 2-2 + 2-4b + 8 ≥ 0, that is, A2 + B2 + B2 ≥ 16. When - A2 > 2, that is, that is, a ＜ - A2 ＞ - 2, that is, that is a \\\\\\\\\\\\2, that is - 4 ≤ a ≤ 4 When (- 2) 2-2a + B-2 ≤ 0 or 22 + 2A + B-2 ≤ 0, that is - 2A + B + 2 ≤ 0 or 2A + B + 2 ≤ 0, A2 + B2 ≥ 45. The minimum value of A2 + B2 is 45
If the function f (x) is even, then the symmetry axis of F (x-1) is? A.x axis b.y axis c.x = 3 D.X = 1
Select d: x = 1
F (x-1) is to shift the original function one unit to the right
Let real numbers a and B make equation X4 + AX3 + bx2 + ax + 1 = 0, and find the minimum value of A2 + B2
Let t = x + 1 x, then T2 + at + B-2 = 0, and t ≥ 2. Let g (T) = T2 + at + at + B-2, (124 ≥ 2). Let g (T) g (t (T) = T2 + 2 + 2 + 2-2 + 2-4b + 8 ≥ 0, that is, A2 + B2 + B2 ≥ 16. When - A2 > 2, that is, that is, a ＜ - A2 ＞ - 2, that is, that is a \\\\\\\\\\\\2, that is - 4 ≤ a ≤ 4 When (- 2) 2-2a + B-2 ≤ 0 or 22 + 2A + B-2 ≤ 0, that is - 2A + B + 2 ≤ 0 or 2A + B + 2 ≤ 0, A2 + B2 ≥ 45. The minimum value of A2 + B2 is 45
The known function f (x) = A / 2-A / 2cos (2 ω + 2 φ) (a > 0, ω > 0,00,0)
A / 2 = 2, a = 4 or a / 2 - (- A / 2) = 2, a = 2 (because f (1) = 2, and 00,0
It is known that a and B are real numbers, and the system of equations y = x3 − AX2 − bxy = ax + B with respect to X and y has integer solutions (x, y). Find the relationship satisfied by a and B
Substituting y = ax + B into y = x3-ax2-bx, eliminating a and B, we get y = X3 XY, then (x + 1) y = X3, if x + 1 = 0, that is, x = - 1, then the left side of the above formula is 0, and the right side is - 1, which is impossible, so x + 1 ≠ 0, then y = x3x + 1 = x2-x + 1-1x + 1 ∵ x, y are all integers ∵ x + 1 = ± 1, that is, x = - 2 or x = 0 ∵ y = 8 or y = 0, so x = - 2
If the function y = f (x) satisfies f (a + x) = f (b-X), then y = the axis of symmetry of F (x)