It is known that the quadratic function f (x) = ax & # 178; + BX (x ≠ 0) satisfies f (- x + 5) = f (x-3), and the equation f (x) = x has equal roots. ① find the analytic expression of the function. ⑥ when x [- 1 / 2,1], the image of the function f (x) is always below the function y = 2x + m, and find the range of M,

It is known that the quadratic function f (x) = ax & # 178; + BX (x ≠ 0) satisfies f (- x + 5) = f (x-3), and the equation f (x) = x has equal roots. ① find the analytic expression of the function. ⑥ when x [- 1 / 2,1], the image of the function f (x) is always below the function y = 2x + m, and find the range of M,

It is known that the quadratic function f (x) = ax & # 178; + BX + C satisfies the following conditions: F (2) = f (0) = 0, and the equation f (x) has two equal roots
Finding the analytic expression of F (x)
A:
f(x)=ax^2+bx+c
f(2)=4a+2b+c=0
f(0)=0+0+c=0
The solution is: C = 0, B = - 2A
So:
f(x)=ax^2-2ax
I don't know what the equation of X is, please add
f（2）=f(0)=0,
Find the values of a and B such that the solution of the inequality ax & # 178; + BX + A & # 178; - 1 ≤ 0 is - 1 ≤ x ≤ 2
That is - 1 and 2 are the roots of the equation AX & # 178; + BX + A & # 178; - 1 = 0
So - 1 + 2 = - B / A
-1×2=(a²-1)/a
a²+2a-1=0
a=-1±√2
The inequality is less than, from the solution set
A>0
So a = - 1 + √ 2
b=-a=1-√2
If x = - 1-ay is substituted into the second B (- 1-ay) - 2Y + 1 = 0 - b-aby-2y + 1 = 0 aby + 2Y = 1-B (AB + 2) y = 1-B with or without array solution, then the coefficients and constants of Y are equal to 0, that is, 0 * y = 0
Given the complete set u = {(x, y) | x ∈ R, y ∈ r}, a = {(x, y) | (Y-3) / (X-2) = 1, X ∈ R, y ∈ r}, B = {(x, y) | y ≠ x = 1, X ∈ R, y ∈ r}, find CUA ∩ cub
Y ≠ x + 1, not y ≠ x = 1
A = {(x, y) | (Y-3) / (X-2) = 1, X ∈ R, y ∈ r} = {(x, y) | y = x + 1, X ≠ 2} B = {(x, y) | y ≠ x + 1}, then a ∪ B = {(x, y) | y = x + 1, X ≠ 2 or Y ≠ x + 1} = {(x, y) | (x, y) ≠ (2,3)} CUA ∩ cub = Cu (a ∪ b) = {(2,3)}
If the solutions of the equations ax by = 6 2x + 3Y = 7 and 4x-5y = 3 ax + by = 2 about x y are the same, find a of B
If ax by = 6, 2x + 3Y = 7 and 4x-5y = 3, ax + by = 2 have the same solution, then 2x + 3Y = 7. (1) 4x-5y = 3. (2) (1) * 2 - (2), 11y = 11y = 1 is substituted into (1) 2X + 3 * 1 = 72x = 4x = 2, x = 2, y = 1 is substituted into ax by = 6, ax + by = 2, 2a-b = 6. (3) 2A + B = 2. (4) (3) + (4), 4A = 8A = 2 is substituted into (3) 2 * 2
First, the two equations containing X and y are combined into a system of equations to obtain the value of X and y, and then the value of X and Y is brought into the two equations containing a and B to obtain ab
Elementary three mathematics binary linear equation (formula method)
It is known that there are two unequal real roots for the square + 2 (k-1) of X + K and the Square-1 = 0 of X
(1) Find the value range of real number K
(2) Could 0 be another root of the equation? If not, ask for another root. If not, please explain the reason
Detailed requirements are strongly required
3Q
First of all, this is about the problem of quadratic equation with one variable
x^2+2(k-1)x+k^2-1=0
1）
If there are two unequal real roots, then:
△=4(k-1)^2-4(k^2-1)＞0
4k^2-8k+4-4k^2+4＞0
8k＜8
k＜1
2）
When x = 0
0+0+k^2-1=0
k^2=1
K = 1 or K = - 1
That is: when k = - 1, x = 0 is a root of the equation
The equation is as follows:
x^2-4x=0
x(x-4)=0
X = 0 or x = 4
The other root of the equation is x = 4
1。
X^2+2(k-1)X+k^2-1=0
Discriminant = 4 (k-1) ^ 2-4 (k ^ 2-1)
=4(k^2-2k+1)-4k^2+4
=8-8k>0
Get K0
8-8k>0
... unfold
(1) Because there are two real equations,
^ 2: [K (2)] - 1 * - 1
(2k-2)^2-4k^2+4>0
4k^2-8k+4-4k^2+4>0
8-8k>0
K
Use the complete square formula to calculate the second power of 895 = the second power of 109 = the second power of 14.5 = (complete solution process)
895²=(900-5)²=900²-2*5*900+25=810000-9000+25=801025109²=(110-1)²=110²-2*1*110+1=12100-220+1=1188114.5²= (14+0.5)²=14²+2*0.5*14+0.5²=196+14+0.25=210.25
Round up.
The general steps of solving linear equation of one variable include removing denominator, removing bracket, moving term, merging similar term, and unifying coefficient
The steps to solve the integral equation of one variable are as follows: 1. Remove the brackets if there are brackets. When removing the brackets, the positive sign is in front of the brackets. After removing the brackets, all items in the brackets will remain unchanged. After removing the brackets, all items in the brackets will be changed. 2. Move the item with unknown number to the left of the equal sign, and the constant item to the equal sign
Let u = {1,2,3,4,5,6,7,8,9} if a ∩ B = {3}, a ∩ (cub) = {1,5,7}, (CUA) ∩ (cub) = {9}, find a, B
Yes
A={1,3,5,7}
B={2,3,4,6,8}
From a ∩ B = {3}, it is known that there are 3 in a and B, and a = {3
B={3……
From a ∩ (cub) = {1,5,7}, we get that there are 157 in a
Guess a = {1,3,5,7 } B={3……
CuB={1,5,7…… }
From (CUA) ∩ (cub) = {9}, we know that there are no 9 in a and B, that is, there are 9 in cuab
Then a = {1,3,5,7 } B={3…… }
Because CUA + a = u cub + B = u
So we can guess that CUA = {2,4,6,8,9}
Cub = {1,5,7,9} (because 157 and 246 in AB must be staggered, respectively in the original set and complementary set)
Checking calculation
A={1,3,5,7}
B={2,3,4,6,8}
CuA={2,4,6,8,9}
CuB={1,5,7,9}
To tell you the truth, this thing is not much technical content, there is no way to give you very rigorous to prove
Given that x = 1 / 2, y = - 1 is the solution of the system ax-3y = 5, 2x-by = 1, then the square of A-B =?
AX-3Y=5,
2X-BY=1
A/2+3=5
1+B=1
B=0
A=4
Square of a - square of B = 16
Sixteen