When x = 2, the value of PX & sup3; + QX + 1 is 2005 How much does this formula equal

When x = 2, the value of PX & sup3; + QX + 1 is 2005 How much does this formula equal

When x = 2,
simple form
=8p+2q+1
=2005
So: 8p + 2q = 2004
When x = - 2,
simple form
=-8x-2q+1
=-（8x+2q）+1
=-2004+1
=-2003

When x = 2, the value of the integral PX & # 179; + QX + 1 is equal to 102, then when x = - 2, the integral PX & # 179; + QX + 1=____
Is it - 102? I'm not sure

When x = 2, the integer PX & # 179; + QX + 1 is equal to 102,8p + 2q + 1 = 102; 8p + 2q = 101; then when x = - 2, the integer PX & # 179; + QX + 1 = - 8p-2q + 1 = - (8p + 2q) + 1 = - 101 + 1 = - 100; Hello, I'm glad to answer for you, skyhunter 002 will answer for you. If you don't understand this question, you can ask

When x = 2, the value of the integer PX3 + QX + 1 is equal to 2002, then when x = - 2, the value of the integer PX3 + QX + 1 is ()
A. 2001B. -2001C. 2000D. -2000

Substituting x = 2 into PX3 + QX + 1 = 2002, we get 23p + 2q + 1 = 2002, that is, 23p + 2q = 2001, when x = - 2, PX3 + QX + 1 = - 23p-2q + 1, = - (23p + 2q) + 1, = - 2001 + 1, = - 2000

When x is equal to 1, the value of cube + QX + 1 of inverse PX is 2010, and it is evaluated when x is equal to - 1

The specific steps of negative 2008 are as follows: when x is equal to 1, the cube of P is equal to 2009-q; when x is equal to negative 1, the cube of P is equal to 2009-q, which is substituted into it to become the cube of negative p - Q + 1 becomes - (2009-q) - Q + 1, which is equal to - 2008

Given that f (x) = (PX ^ 2 + 2) / (3x + Q) is an odd function and f (2) = 5 / 3, try to find the value of P and Q

Since it is an odd function, f (x) = - f (x), f (2) = - f (- 2)
Two equations can be obtained
f(2)=(4p+2)/(6+q)=5/3
f(-2)=(4p+2)/(-6+q)=-f(2)=-5/3
The solution of the equations is: P = 2, q = 0

A necessary and sufficient condition for Ax & sup2 + 2x + 1 = 0 (a ≠ 0) to have at least one positive root

If there is a root, then 4-4a > = 0
a

If two univariate quadratic equations a 2x 2 + AX-1 = 0 and X 2-ax-a 2 = 0 about X have common solutions, all the values of a are obtained

Let the common root of the equation be B, then substitute it into the above two equations: (AB) 2 + AB-1 = 0, ①, b2-ab-a2 = 0, ② add the above two equations: B2 (A2 + 1) - (A2 + 1) = 0, ■ (b2-1) (A2 + 1) = 0, the solution is: B = 1 or - 1; when B = 1, substitute it into the second equation: A2 + A-1 = 0; according to the root formula, a = 1 ± 52 can be obtained; when B = - 1, substitute it into the second equation: a2-a-1 = 0 In conclusion, the value of a is 1 + 52 or 1 − 52

From the four numbers 2.3.4.5, arbitrarily take two numbers P and Q (P is not equal to q) to form the functions y = px-2 and y = x + Q, and make the intersection of the two numbers on the right side of x = 2,
Then (P, q) consists of a 12 pairs, B 6 pairs, C 5 pairs, D 3 pairs,

10.2

If two of the equations x ^ 2-px + q = 0 (P and Q belong to real numbers) are X1 and X2, then the quadratic equation with - X1 and - x2 as roots is?

X1+X2=p,X1X2=q,
So (- x1) + (- x2) = - P, (- x1) (- x2) = q,
So the quadratic equation with - x1, - x2 as roots is x ^ 2 + PX + q = 0

If the positive root of quadratic function x & # 178; - px-q = 0 (P, Q ∈ n) is less than 3, then such quadratic function has? A, 5 B, 6 C, 7 d and 8 d

From the meaning of the title, we get x ^ 2-px-q = (x-m) (x-n), where M