Given the 4kx + 3 (1) under f (x) = 2x + 3 / radical, if the domain of F (x) is r, find out whether the value (2) of real number k has real number k, so that the domain of F (x) is r（ Given f (x) = 2x + 3 / 4kx + 3 under root sign (1) If the domain of F (x) is r, find the value of real number K (2) Is there a real number k such that the domain of F (x) is (- ∞, - 2)? If so, find the value of the real number K. if not, explain the reason,

Given the 4kx + 3 (1) under f (x) = 2x + 3 / radical, if the domain of F (x) is r, find out whether the value (2) of real number k has real number k, so that the domain of F (x) is r（ Given f (x) = 2x + 3 / 4kx + 3 under root sign (1) If the domain of F (x) is r, find the value of real number K (2) Is there a real number k such that the domain of F (x) is (- ∞, - 2)? If so, find the value of the real number K. if not, explain the reason,

(1) The domain of F (x) is 4kx + 3 > 0
If the domain of F (x) is r, then k = 0 must be satisfied
(2)、∵4kx+3>0
∴4kx>-3
Let f (x) be defined as (- infinity, - 2)
It must be K

Given the symmetric axis of quadratic function x = - 2, and the length of the line segment cut by the intersection of image and Y axis (0,1) on X axis is 2 √ 2, the expression is obtained

Answer: let the quadratic function be y = a (x + 2) &# 178; + C point (0,1) to get 4A + C = 1 (1) Let y = a (x + 2) ² + C = 0, and the solution is: x = - 2 ± √ (- C / a), so: x2-x1 = 2 √ (- C / a) = 2 √ 2, so: - C / a = 2, C = - 2A (2) From (1) and (2) solution: a = 1 / 2, C = - 1, so: quadratic function

(Guilin, 2012) as shown in the figure, if the image of function y = AX-1 passes through point (1,2), then the solution set of inequality AX-1 > 1 is

If the image passes through point (1,2), then x = 1, y = 2, y = AX-1, 2 = A-1, a = 3
Substituting a = 3 into AX-1 > 1, 3a-1 > 1, a > 2 / 3 is obtained

Given that f (x) is a quadratic function, and f (0) = 3, f (x + 2) - f (x) = 4x + 2, find f (x)

If the expansion of (x to the second power - 3x + 4) (x to the second power - ax + 1) contains the coefficient - 1 of the quadratic term of X, the value of a is obtained

After expansion, those with X & # 178; points:
x²+4x²+3ax²=（5+3a）x²
（5+3a）=-1
a=-2

If the coefficient of (x2-3x + 4) (x2-ax + 1) is - 1, then the value of a is______ ．

The coefficients of (x2-3x + 4) (x2-ax + 1) = x4-ax3 + x2-3x3 + 3ax2-3a + 4x2-4ax + 4 = x4-ax3-3x3 + (1 + 3A + 4) x2-3a-4ax + 4 ∵ x2 are - 1, ∵ 1 + 3A + 4 = - 1, a = - 2

(1+1/2)x(1+1/4)x(1+1/6)x...x(1+1/10)x(1-1/3)x(1-1/5)x..x(1-1/9)

(1+1/2)x(1-1/3)=1
(1+1/4)x(1-1/5)=1
(1+1/2)x(1+1/4)x(1+1/6)x...x(1+1/10)x(1-1/3)x(1-1/5)x..x(1-1/9)=1

(1-1/2)x(1-1/3)x(1-1/4)x...x(1-1/9)x(1-1/10)

(1-1/2)x(1-1/3)x(1-1/4)x.x(1-1/99)x(1-1/100)
=1 / 2 * 2 / 3 * 3 / 4 *. * 99 / 100 (all internal offsets)
=1/100

1/x+1+1/x+10=1/x+2+1/x+9

1/(x+1)+1(/x+10)=1/(x+2)+1/(x+9)
1/(x+1)-1(/x+9)=1/(x+2)-1/(x+10)
.
(x+1)(x+9)=(x+2)(x+10)
x=-5.5

How to solve the equation {x + 3Y = 120 3x-y = 8

x=14.4
y=35.2