Ax + BX = 3, ax ^ 2 + by ^ 2 = 7, ax ^ 3 + by ^ 3 = 16, ax ^ 4 + by ^ 4 = 42, find ax ^ 5 + BX ^ 5

Ax + BX = 3, ax ^ 2 + by ^ 2 = 7, ax ^ 3 + by ^ 3 = 16, ax ^ 4 + by ^ 4 = 42, find ax ^ 5 + BX ^ 5

Ax ^ 2 + by ^ 2 = 7 (AX ^ 2 + by ^ 2) (x + y) = 7 (x + y) ax ^ 3 + by ^ 3 + ax ^ 2Y + bxy ^ 2 = 7 (x + y) ax ^ 3 + by ^ 3 + XY (AX + by) = 7 (x + y) substitute ax + by = 3, ax ^ 3 + by ^ 3 = 16 into the above equation 16 + 3xy = 7 (x + y) ax ^ 3 + by ^ 3 = 16 (AX ^ 3 + by ^ 3) (x + y) = 16 (x + y) =
If a, B, x, y satisfy ax + by = 3, ax ^ 2 + by ^ 2 = 7, ax ^ 3 + by ^ 3 = 16, ax ^ 4 + by ^ 4 = 42, find the value of ax ^ 5 + by ^ 5
I will
F（n）=ax^n+by^n
x. Y is TT = Pt + Q
F（n)=pF(n-1)+qF(n-2)
F(3)=pF(2)+qF(1)
F(4)=pF(3)+qF(2)
p=-14
q=38
F(5)=20
Is the premise of using acute trigonometric function in right triangle?
I know that any angle or other graph can be transformed into a triangle to use the acute angle trigonometric function, but I want to know whether the triangle divided from any angle or other graph can only use the acute angle triangle if it is a right triangle?
No~
You should be a junior high school student? Junior high school will have the concept of acute angle trigonometric function High school has expanded to any corner
And this is just a trigonometric function, the object is arbitrary angle, of course, you can cut the triangle into right triangles to calculate, but generally no one will do this
Is there anything you don't understand?
37 / 39 + 7 / 25_ 37 of 39 + 8 of 25 = (calculated by simple method)
=37/39 -37/39 +7/25+8/25
=(37/39 -37/39) +(7/25+8/25)
=0+15/25
=3/5
Finding the square of function range y = x-4x-5 (x > 1)
y=x^2-4x+4-9=(x-2)^2-9
Because when x > 1 and x = 2 (X-2) = 0, the range of Y is Y > = - 9
When the real number a takes what value, the equation LG (x-1) + LG (3-x) = LG (1-ax) has one solution, two solutions and no solution?
x-1>0-->x>1
3-x>0--> x 1 ax a a=0 ,-8
1)delta
Domain of definition
x-1>0
3-x>0
1-ax>0
The above two formulas give 1