The function f (x) satisfies f (2x) = 5x / (xsquare - x + 1). The analytic expression and domain of F (x + 1) are obtained

The function f (x) satisfies f (2x) = 5x / (xsquare - x + 1). The analytic expression and domain of F (x + 1) are obtained

Let t = 2x
x=t/2
f(t)=10t/(t^2-2t+1)
Let f (T + 1) = 10t + 10 / (T ^ 2)
T is not equal to 0
f(x+1)=(10x+10)/(x^2)
Domain: X is not equal to 0
You can ask me if you don't understand
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In the triangle ABC, ab = BC = 3, AC = 4, let o be the interior of the triangle ABC if the vector Ao = m, the vector AB + N, the vector AC
Find M: n=
If O is the heart, then it is easy to calculate that the radius of the inscribed circle of the triangle is 1, that is, the distance from O to the three sides is 1. Make the vertical lines AB and AC respectively through O, and the vertical feet are m and N, then | am | = | an | = 1
(vector arrow I omitted)
AO=AM+AN=1/3AB+1/4AC=1/3AB+1/4(AB+BC)=7/12AC+1/4BC
So m + n = 7 / 12 + 1 / 4 = 5 / 6
2 one variable quadratic equation discriminant problem! Will come!
1. When k takes what value, the equation x & sup2; - K (x-1) + x = 0 has two equal real roots
2. Given the equation x & sup2; + 2kx + (K-2) & sup2; = x about X, when k takes what value, equation (1) has two unequal real roots (2) has two equal real roots (3) has no roots
1. From the meaning of the question: two equal roots of real numbers, that is △ = 0
△=k²-4×1×1=0 k²=4 k=±2
A: when k = ± 2, there are two equal real roots
2、
(1) ∵ two unequal real roots △ 0
∴（2k）²-4×1×（k-2）²＞0
4k²-4k²+16k-16＞0
16k-16＞0
k＞1
Answer: when k > 1, there are two unequal real roots
(2) ∵ two equal real roots △ = 0
∴（2k）²-4×1×（k-2）²=0
4k²-4k²+16k-16=0
16k-16=0
K=1
Answer: when k = 1, there are two unequal real roots
（3）∵△＜0
∴（2k）²-4×1×（k-2）²＜0
4k²-4k²+16k-16＜0
16k-16＜0
k＜1
When k > 1, there are two unequal real roots
1。 (- K + 1) '2-4k = 0 K'2-6k + 1 = 0 K = 3 + (-) 2 * radical 2
2。 4k'2-4*(k-2)'2=4*2*(2k-2)=16(k-1)
(1),k-1>0 k>1
(2),k-1=0 k=1
(3),k-1
For example, 2x-6 equals 3x + 5,
2x-6=3x+5
2x-3x=6+5
-x=11
x=-11
This is the step to solve the problem. You should study it yourself
2x-6=3x+5 2x-6-3x-5=0 2x-3x-6-5=0 -x=11 x=-11
To move items, we need to change the sign - 6-5 = 3x-2x, so x = - 11
2x-6=3x+5
2x-3x=5+6
-x=11
x=-11
Merge congeners
a^4+5a^2b^2+9b^4 4a^4+24a^2b^2+25b^4
a^4+5a^2b^2+9b^4
=a^4+6a²b²+9b^4-a²b²
=(a²+3b²)²-a²b²
=(a²+ab+3b²)(a²-ab+3b²)
4a^4+24a^2b^2+25b^4
=4a^4+24a²b²+36b^4-11b^4
=(2a²+6b²)²-11b^4
=[2a²+(6+√11)b²][2a²+(6-√11)b²]
The known function f (x) = 2x (0 ＜ = x ＜ = 1) f (x) = - 2 / 5x + 12 / 5 (1 ＜ x ＜ = 5)
(1) If there are two intersections between the image of function y = f (x) and the line kx-y-k + 1, the value range of real number k is obtained
(2) Try to find the range of function g (x) = XF (x)
F (x) = {2x (0 ＜ = x ＜ = 1); {- 2 / 5x + 12 / 5 (1 ＜ x ＜ = 5). Its image is a broken line OAB, where o is the origin, a (1,2), B (5,2 / 5). (1) the straight line L: kx-y-k + 1 = 0 passes through the point C (1,1). When it turns from CB to CO, it has exactly two intersections with the image of y = f (x)
Given that vectors a and B are not collinear, if vector AB = λ 1A + B and vector AC = a + λ 2B, then ABC three points are collinear
Given that vectors a and B are not collinear, if vector AB = λ 1A + B, vector AC = a + λ 2B, and three points a, B and C are collinear, then real numbers λ 1 and λ 2 must satisfy λ 1, λ 2 = 1 answer: λ 1A + B = K (a + λ 2b) ∧ λ 1A + B = Ka + K λ 2B ∧ λ 1 = k, 1 = λ 2K
The quadratic equation of one variable x ^ 2-6x-k ^ 2 = 0 is known
K is a constant
There are two real roots that don't want to wait
Let X1 and X2 be the two real roots of the equation, and X1 + 2x2 = 14, try to find the two real roots of the two equations
1) ∵ Δ = 36 + 4K & # 178; > 0, the equation has two unequal real roots
2) ∵ x1, X2 are the two real roots of the equation
It is concluded from the Veda theorem that:
x1+x2=6,
And X1 + 2x2 = 14
By solving the equations, X1 = - 2, X2 = 8
1. It is proved that: △ = 36-4 × 1 × (- K & # 178;) = 36 + 4K & # 178; > 0
There are two real roots that don't want to wait
2. The relationship between root and coefficient, X1 + x2 = 6
∵x1+2x2=14
∴x1=-2 x2=8
Is 2x-7 = 3x-11 a linear equation with one variable
Look at the definition of once a dollar in mathematics
One yuan: only one unknown x
Once: the highest power of X is "1"
So, it's a linear equation of one variable
Factorization 4A ^ 3B ^ - 12a ^ 2B=
4a^3b^2-12a^2b
=4a²b(ab-3)
Original formula = 4A & # 178; B (ab-3)