If f (x) is a linear function, 2f (x-1) + 3f (x + 1) = 5x + 6, the analytic expression of F (x) can be obtained Let f (x) be f (x) = KX + B, and then?

If f (x) is a linear function, 2f (x-1) + 3f (x + 1) = 5x + 6, the analytic expression of F (x) can be obtained Let f (x) be f (x) = KX + B, and then?

Answer: Let f (x) = KX + B, then substitute 2x-1 and X + 1 to get: F (2x-1) = K (2x-1) + B = 2kx + b-kf (x + 1) = K (x + 1) + B = KX + B + K into 2F (2x-1) + 3F (x + 1) = 5x + 6 to get: 4kx + 2b-2k + 3kx + 3B + 3K = 5x + 67kx + 5B + k = 5x + 6, so: 7K = 55B + k = 6, the solution is: k = 5 / 7, B = 37 / 35, so: F (x) = 5x /
It is known that f (x) is a linear function,
Let f (x) = ax + B
Then: F (x-1) = a (x-1) + B = ax-a + B
f(x+1)=a(x+1)+b=ax+a+b
It is known that 2F (x-1) + 3f (x + 1) = 5x + 6
So: 2 (ax-a + b) + 3 (AX + A + b) = 5x + 6
2ax-2a+2b+3ax+3a+3b=5x+6
5ax+a+5b=5x+6
In the triangle ABC, ab = 3, AC = 4, BC = 5, O is the heart of the triangle ABC, and the vector Ao = mAb + NBC (AO, AB, AC are all vectors), then M + n =?
Let | ab | = C, | AC | = B, | BC | = a, then
A * vector OA + b * vector ob + C * vector OC = 0, (1)
Also: vector ob = vector (oa-ba), vector OC = vector (oa-ca), vector AC = vector (AB + BC),
From (1), we can get that,
A * vector OA + b * vector (oa-ba) + C * vector (oa-ca) = 0,
A * vector OA + b * vector (oa-ba) + C * vector (OA + AB + BC) = 0,
(a + B + C) * vector OA = - B * vector ab-c * vector (AB + BC),
Vector OA = [- (B + C) * vector ab-c * vector BC] / (a + B + C). (2)
Also: vector Ao = m vector AB + n vector BC. (3)
By comparing the coefficients of equations (2) and (3), we can get that,
M=(b+c)/(a+b+c)=(3+4)/12=7/12; N=c/(a+b+c)=5/12
∴M+N=1
What is the discriminant of quadratic equation of one variable?
ax²+bx+c=0
The discriminant is B & # 178; - 4ac
A solution to the equation of one variable a problem (to brackets)!
2（10-0.5y）=-（1.5y+2)
Write down the process of removing brackets
2*10-2*0.5y=-1.5y-2
20-y=-1.5y-2
-y+1.5y=-2-20
0.5y=-22
y=-44
I'm not good at math either (1.5y + 2). Is there a minus sign in front of it?
20-y=-1.5y-2
0.5y=-22
y=-44
The original form is changed into
20-Y=-1.5Y-2
Transfer of items,
0.5Y=-22
So y = - 44
Original equation = 20-y = - 1.5y-2
-y+1.5y=-2-20
0.5y=-22
y=-44
2*10-2*0.5y=-1.5y-2
20-y=-1.5y-2
-y+1.5y=-2-20
0.5y=-22
y=-44
20-y=-1.5y-2
0.5y=-22
y=-44
20－y=1.5y+2
20-2=2.5y
18=2.5y
18/2.5=y
7.2=y
20-y=-1.5y-2
0.5y=-22
y=-44
Factorization of (a ^ 2 + B ^ 2) ^ 2-4a ^ 2B ^ 2
Given the function f (X-2) = 2x ^ 2-5x + 1, find f (x)
Let t = X-2, x = t + 2, then
f(t)=2(t+2)^2-5(t+2)+1
That is, f (T) = 2T ^ 2-2t-1
∴f(x)=2x^2-2x-1
Tell you a way
Let X-2 = t
x=t+2
f(t)=2(t+2)^2-5(t+2)+1
=2(t^2+4t+4)-5t-10+1
=2t^2+8t+8-5t-9
=2t^2+3t-1
Then we can get it by replacing T with X (note that the direct substitution is not replaced by X-2 = t)
f(x)=2x^2+3x-1
Let X-2 = K
So, x = K + 2
So f (k) = 2 (K + 2) ^ 2-5 (K + 2) + 1 = 2 (k ^ 2 + 4K + 4) - 5k-10 + 1 = 2K ^ 2 + 3K-1
So f (x) = 2x ^ 2 + 3x-1
By substitution
Let x = t + 2 and substitute f (X-2) = 2x ^ 2-5x + 1 to get f (T) = 2 (T + 2) ^ 2-5 (T + 2) + 1
It is concluded that f (T) = 2T ^ 2 + 3t-1
That is, f (x) = 2x ^ 2 + 3x-1
Matching method
f(x-2)=2（x^2-4x+4)+3（x-2)-1
=2(x-2)^2+3(x-2)-1
That is, f (x) = 2x ^ 2 + 3x-1
In triangle ABC, ab = 2, AC = 3, x + 2Y = 1, O is the heart, vector Ao = x, vector AB + y, vector AC, find cosa
If a quadratic equation with one variable has two equal virtual roots, is its discriminant also equal to 0
Yes, it's just that the coefficient is imaginary
If the coefficients are all real numbers, it is impossible to have equal imaginary roots
What is the basis of the term shift of the linear equation of one variable?
If both sides of the equation add or subtract the same number, the equation still holds
4A ^ 4 + 12a ^ 2B ^ 2 + 9b ^ 2 factorization