If the function f (x) defined on R holds f (x1 + x2) = f (x1) + F (x2) - 1 for any x1, X2 ∈ R, and if x ＞ 0, f (x) ＞ 1 (1) To prove that f (x1) - 1 is an odd function (2) It is proved that f (x) is an increasing function on R (3) If f (4) = 5, solve the inequality f (3m to the second power - m-2) < 3

If the function f (x) defined on R holds f (x1 + x2) = f (x1) + F (x2) - 1 for any x1, X2 ∈ R, and if x ＞ 0, f (x) ＞ 1 (1) To prove that f (x1) - 1 is an odd function (2) It is proved that f (x) is an increasing function on R (3) If f (4) = 5, solve the inequality f (3m to the second power - m-2) < 3

1 Let x2 = 0; when X1 > 0, f (x1) = f (x1) + F (0) - 1; from this, we can get that f (0) = 1; then let x2 = - x1, f (0) = f (x1) + F (- x1) - 1 is reduced to f (x1) - 1 = - f (- x1) + 1; thus, we can prove that f (x1) - 1 is an odd function
In triangle ABC, vector AB = (1,2), vector AC = (4x, 3x), where x > 0, the area of triangle ABC is 5 / 4, what is the value of real number x
process
|AB|=√(1+2)=√5,|AC|=√((4x)+(3x))=5x ∴cosBAC=AB*AC/|AB|*|AC|=(4x+6x)/(√5×5x)=2/√5 ∴sinBAC=√(1-cosBAC)=1/√5 ∴S△ABC=(1/2)|AB|×|AC|×sinBAC=(1/2)×√5×5x×1/√5=5x/2=5/4 ∴x=1/2
The discriminant problem of the root of quadratic equation with one variable
Ax ^ 2-4x + A + 3 > 0 is constant, and the value range of a is obtained
When a is not equal to 0, a > 0 and △ = 16-4a (a + 3)
Another problem: given that the range of F (x) = log (x ^ 2-2ax + 4-3a) is r, find the range of A
There is a step in the answer: x ^ 2-2ax + 4-3a = 0 △ > = 0. Why not
Ax ^ 2-4x + A + 3 > 0 is constant, that is, the image of the quadratic function f (x) = ax & sup2; - 4x + A + 3 is always above the x-axis, that is, the opening of the parabola is upward and there is no intersection point with the x-axis, so a > 0 has no intersection point with the x-axis, so when △ = 0, the parabola has an intersection point with the x-axis, and it is impossible to ensure that the throwing lines are all above the x-axis, that is
From the combination of number and shape, a > 0 means that the opening is upward, and △
Is 1 / 2x-5 = 3 a linear equation with one variable?
The following equation is a linear equation of one variable, and the result is ()
A.x+y=1
B.x²+5x=3
C.3x+7=16
D.1/2x-5=3
I chose C D
As a result, the teacher hit X
The correct answer is C. I don't know why D is not
D is a fractional equation written as a product of 1 times (2x) to the negative power
Factorization: (a + b) &# 178; - (a-2b) &# 178; urgent
Original formula: = A & # 178; + 2Ab + B & # 178; - A & # 178; + 4ab-4b & # 178; = 6ab-3b & # 178; = 3B (2a-b) I'm very happy to answer for you, like the rising sun, I have a small request, your affirmation is the driving force of my answer. Your support encourages me to move on
=a²+2ab+b²-(a²-4ab+4b²)
=a²+2ab+b²-a²+4ab-4b²
=6ab-3b²
Let f (x) be a function over the domain of definition R. if f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2) has f (x1 + x2) + F (x1-x2) = 2F (x1) f (x2) for any x 1 and x 2, then f (x) is odd
Let x2 = 0: F (x1) + F (x1) = 2F (x1) f (0)
Since it holds for any formula over x1, we obtain that f (0) = 1
Let X1 = 0, then f (x2) + F (- x2) = 2F (0) f (x2) = 2F (x2)
∴f(－x2)＝f(x2)
F (x) is an even function
Given vector a = (x, 1), B = (3, 6), a ‖ B, then the value of real number x is ()
A. 12B. -2C. 2D. -12
∵ vector a = (x, 1), B = (3, 6), a ∥ B, ∥ has non-zero real number μ, so that B = μ a, 3 = μ X6 = μ, and the solution x = 12
Discriminant of the root of quadratic equation with one variable
1. Prove that the quadratic equation x ^ 2 + (a + 1) x + 2 (A-2) = 0 has two unequal real roots
2. Given that the equation x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots, try to judge whether the straight line y = (2m-3) x-4m-7 passes through point a (- 2,4)? Explain the reason
3. Determine the root of the equation (x-a) (X-2) = 0, and explain the reason
1. Proof: from the question, according to b-4ac, we get
(a + 1) - 8 (A-2)
= square a + 2A + 1-8a + 16
= square A-6A + 17
= (A-3) + 8
Because no matter what position a takes, the above formula is always greater than 0,
Therefore, the equation must have two unequal real roots
2. According to the original equation, it can be obtained from b-4ac
m＝7/4
Substituting it into the linear equation, we get
y＝1/2x-14.
Therefore, it does not go through (- 2,4)
3. According to the formula b-4ac, it can be changed into formula (A-2)
When a = 2, the equation has two equal real roots,
When a is not equal to 2, the equation must have two unequal real roots
1.(a+1)^2-4*1*2(a-2)=a^2-6a+17=(a-3)^2+8>0,
So two real roots are not necessarily equal
2. X ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots,
So (2m + 1) ^ 2-4 * 1 * (m ^ 2 + 2) = 0, the solution is m = 7 / 4
The straight line y = (2m-3) x-4m-7 can be reduced to y = x / 2-14,
4 is not equal to - 2 / 2-14
To begin with
1.(a+1)^2-4*1*2(a-2)=a^2-6a+17=(a-3)^2+8>0,
So there must be two unequal real roots
2. X ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots,
So (2m + 1) ^ 2-4 * 1 * (m ^ 2 + 2) = 0, the solution is m = 7 / 4
The straight line y = (2m-3) x-4m-7 can be reduced to y = x / 2-14,
4 is not equal to - 2 / 2-14
So it's just a (- 2,4)
3. When a = 2, there is one X = 2; when a is not equal to 2, there are two, X1 = a, X2 = 2
(x-a)(x-2)=0,
(x-a) = 0 or (X-2) = 0 = retract
1. Prove that the quadratic equation x ^ 2 + (a + 1) x + 2 (A-2) = 0 has two unequal real roots
b^2-4ac=(a+1)^2-4*2(a-2)=a^2+2a+1-4a+8=(a-1)^2+9>0
There must be two unequal real roots
2. Given that the equation x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots, try to judge whether the straight line y = (2m-3) x-4m-7 passes through point a (- 2, 4)? Explain why 0
There must be two unequal real roots
2. Given that the equation x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots, try to judge whether the straight line y = (2m-3) x-4m-7 passes through point a (- 2, 4)? Give reasons
The equation x ^ 2 + (2m + 1) x + m ^ 2 + 2 = 0 has two equal real roots
(2m+1)^2-4(m^2+2)=0
m=7/4
The linear y = (2m-3) x-4m-7
y=1/2x-14
Point a (- 2,4)
It doesn't match
So it's no longer straight
3. Determine the root of the equation (x-a) (X-2) = 0, and explain the reason
X = 2 or x = a
The first question: using △ = B ^ 2-4ac = (a + 1) ^ 2-8 (A-2) = a ^ 2-6a + 17 = (A-3) ^ 2 + 8 is always greater than zero, so there are two unequal real roots
1。 The discriminant of root = (a + 1) ^ 2-4 * 1 * 2 (A-2) = a ^ 2-6a + 17 = (A-3) ^ 2 + 8 > 0
So there must be two unequal real roots
2。 Because there are two equal real roots
So the discriminant of root = 0, so the discriminant of root = (2m + 1) ^ 2-4 * 1 * (m ^ 2 + 2) = 0
4m-7=0,m=7/4
The straight line y = (2m-3) x-4m-7 passes through point a (- 2, 4), which is equivalent to 4 = - 8m-1, M = - 5 / 80
So there must be two unequal real roots
2。 Because there are two equal real roots
So the discriminant of root = 0, so the discriminant of root = (2m + 1) ^ 2-4 * 1 * (m ^ 2 + 2) = 0
4m-7=0,m=7/4
The straight line y = (2m-3) x-4m-7 passes through point a (- 2,4), which is equivalent to 4 = - 8m-1, M = - 5 / 8
So the straight line y = (2m-3) x-4m-7 is no more than point a (- 2,4).
3。 When a = 2, there are equal roots. Otherwise, there are two unequal roots: A, 2
*********************
For Ax ^ 2 + BX + C = 0;
The discriminant of root = B ^ 2-4 * a * C ﹣ retract
Is 2x & # 178; + 1-3 = 2 (x-1) a linear equation with one variable?
no
If the polynomial 2x & sup2; - 5xy-3y & sup2; + 3x + 5Y + K can be decomposed into the product of two first-order factors, then K=
2X & # 178; - 5xy-3y & # 178; + 3x + 5Y + K ∵ can be decomposed into the product of two linear expressions
=(2x+y+a)(x-3y+b)
=2x²-5xy-3y²+(（2b+a)x+(b-3a)y+ab)
∵（2b+a)=3
(B-3A) = 5: a = - 1, B = 2
∴k=ab=-2