Given that the function f (x) = (the square of AX + 1) divided by (BX + C) (a, B, C are integers) is odd, and f (1) = 2, f (2) is less than 3, find the value of a, B, C Help get rid of it

F (x) = ax ^ 2 + 1 / BX + C (a, B, C are integers) is an odd function, monotonically increasing in [1, positive infinity], f (1) = 2, f (2)

The general term formula is the sum of the first n terms of the sequence with an equal to 3 to the nth power plus 2n plus 1

An = 3 ^ n + 2n + 1 Let BN = 3 ^ n (arithmetic sequence) B1 = 3Q = 3CN = 2n + 1 (arithmetic sequence) C1 = 3D = 2 the sum of the first n terms of the sequence: an = BN + CN = 3 * (3 ^ n-1) / (3-1) + (1 / 2) * (3 + 2n + 1) n = (3 / 2) * (3 ^ n-1) + (n + 2) n = (1 / 2) * 3 ^ (n + 1) - (3 / 2) + n ^ 2 + 2n = (1 / 2) * 3 ^ (n + 1) + n ^ 2 + 2n - (3 / 2) (n

PA, Pb and PC are three rays starting from point P. the angle between each two rays is 60 degrees. Then the cosine of the angle between PC and PAB is ()
A. 12B. 22C. 33D. 63

Take any point D in PC and make do ⊥ plane APB, then ∠ DPO is the angle between PC and plane PAB; Because do \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\switchto DP O = oppd = 33. That is to say, the cosine of the angle between the line PC and the plane PAB is 33

79 * 42 + 79 + 79 * 57 simple operation

79*42+79+79*57
=79×(42+1+57)
=79×100
=7900
Happy study

Is there anything else in the formula for calculating the area of a parallelogram besides the bottom × height? Is it the product of diagonals
Calculation formula of parallelogram area
Is there anything else besides the bottom x height?
The product of diagonals, right?
How else
thank you

For any quadrilateral, as long as the diagonals are perpendicular to each other, then the area of the quadrilateral is multiplied by the diagonals and divided by two

Solving equation x (x + 7) = 60

X(X+7)=60
x^2+7x-60=0
(x+12)(x-5)=0
x1=-12
x2==5

Given that the circle x * x + y * y + 2x-4y + 1 = 0 is symmetric with respect to the line 2aX by + 2 = 0 (a, B belong to R), then the value range of AB is?
Why can an equivalent line pass through the center of a circle? Is the symmetry of a circle about a line passing through the center of a circle? I don't understand

X & sup2; + Y & sup2; + 2x-4y + 1 = 0, that is, (x + 1) & sup2; + (Y-2) & sup2; = 4
The center of the circle is (- 1,2) and the radius is 2
On the line 2aX by + 2 = 0 (a, B belong to R) symmetry
A straight line passes through the center of a circle
-2a-2b+2=0
a+b=1

If you want the fraction x2 divided by 2x + 1 to be a positive number, then the value range of X is____ .
This is the question of eight fractions... Please reply quickly,
Khamsa Hamida`

x²/(2x+1)>0
Two conditions need to be met
That is, the denominator is greater than 0
The numerator is not zero because the square term is nonnegative
Intrinsic molecule > 0 2x + 1 > 0
x²>0 2x+1>0
Namely
x≠0 x>-1/2
in summary
When x > - 1 / 2 and X ≠ 0, the value of fraction x2 divided by 2x + 1 is positive
Note that you must not forget the numerator! And the denominator cannot be equal to 0, so it is 2x + 1 > 0 x > - 1 / 2
Not equal to!

Given that the function f (x) = 4x ^ 2-2 (P-2) x-2p ^ 2-P + 1 has at least one real number C in the interval [- 1,1], such that f (c) > 0, the value range of real number P is obtained

On the contrary, the function f (x) does not have real number C on [- 1,1], such that f (c) > 0, axis of symmetry ≤ - 1 or ≥ 1, f (- 1) < 0, f (1) < 0, we can get the range of P and then take the complementary set

Given that the function f (x) = (the square of AX + 1) divided by (BX + C) (a, B, C are integers) is odd, and f (1) = 2, f (2) is less than 3, find the value of a, B, C Help get rid of it