The second chapter is quadratic function It is known that the parabola y = ax & # 178; + BX + C intersects with the Y axis at point C, and intersects with the X axis at points a (x1,0), B (x2,0) (x1 < x2), and the ordinate of vertex m is - 4, if X1 and X2 are two roots of the equation x & # 178; - 2 (m-1) x + M & # 178; - 7 = 0, and X1 & # 178; + x2 & # 178; = 10 (1) Find the coordinates of two points a and B; (2) Find the analytic formula of parabola and the coordinates of point C (3) Is there a point P on the parabola so that the area of the triangle PAB is equal to twice the area of the quadrilateral ACMB? If so, find out the coordinates of the qualified p; if not, explain the reason

The second chapter is quadratic function It is known that the parabola y = ax & # 178; + BX + C intersects with the Y axis at point C, and intersects with the X axis at points a (x1,0), B (x2,0) (x1 < x2), and the ordinate of vertex m is - 4, if X1 and X2 are two roots of the equation x & # 178; - 2 (m-1) x + M & # 178; - 7 = 0, and X1 & # 178; + x2 & # 178; = 10 (1) Find the coordinates of two points a and B; (2) Find the analytic formula of parabola and the coordinates of point C (3) Is there a point P on the parabola so that the area of the triangle PAB is equal to twice the area of the quadrilateral ACMB? If so, find out the coordinates of the qualified p; if not, explain the reason

It is suggested that: (1) according to the meaning of the title, we can get X1 + x2 = 2 (m-1), x1 · x2 = M & # 178; - 7, X1 & # 178;; + x2 & # 178;; = 10, we can get m = 2, x1 = - 1, X2 = 3, a (- 1,0), B (3,0); and (2) according to the meaning of the title, we can get A-B + C = 0, 9A + 3B + C = 0, (4ac-b & # 178;) / 4A = - 4, we can get a = 1, B

Junior three mathematics, on the inverse proportion function
The positive half axes of the straight line L, X axis and Y axis intersect at two points a and B respectively, OA = 6 and ob = 8. The moving point P starts from O and moves to point a at a speed of one unit per second. At the same time, the moving point Q starts from point a and moves to point B at two units per second. When one of the points reaches the end point, the other stops moving. Let the moving time be T seconds
When PQ / / ob, is there a point m on the y-axis so that △ MPQ is an isosceles triangle? If so, please write the coordinates of M

Let the coordinates of M be (0, a)
According to the meaning of the title, a coordinates are (6,0), B (0,8), P (T, 0); when PQ / / ob, the abscissa of Q point is t, and Q is on the straight line AB, so the coordinates of Q are (T, - 4 / 3T + 8), and QA = 2T (according to the meaning of the title, the motion speed is two units)
Then according to the distance formula, we can get √ [(T-6) ^ 2 + (- 4 / 3T + 8) ^ 2] = 2T, and we can get t = 30 / 11 (seconds) (t = - 30 rounding off)
Then the coordinates P (30 / 11,0); Q (30 / 11,48 / 11) can be obtained
In this case, PQ = 48 / 11, MP = √ [(30 / 11) ^ 2 + A ^ 2]; MQ = √ [(30 / 11) ^ 2 + (a-48 / 11) ^ 2]
When PQ = MP, △ MPQ is an isosceles triangle, then a = ± 6 √ 39 / 11; that is, the coordinate m (0, ± 6 √ 39 / 11)
When PQ = MQ, △ MPQ is an isosceles triangle, then a = (48 ± 2 √ 351) / 11, that is, the coordinate m (0, (48 ± 2 √ 351) / 11)
When MP = MQ, △ MPQ is an isosceles triangle, then a = 24 / 11, which is the coordinate m (0,24 / 11)
To sum up, there is m on the y-axis, which can make △ MPQ an isosceles triangle. At this time, m coordinates are (0,6 √ 39 / 11) or (0, - 6 √ 39 / 11) or (0, (48 + 2 √ 351) / 11) or (0, (48-2 √ 351) / 11) or (0,24 / 11)

It is right to write in the exercise book that "y = f (x) and y = f (T) are the same function"
But the other question says "f (x), f (T) denote two different functions whose independent variables are x and T"
What's going on? When is the same function and when is not?

The above is to study the relationship between two functions, and the following is to study two functions. They have different meanings and are two different problems

The side length of a square with an area of 19 is x, and X is 19__ Arithmetic square root____ .

Arithmetic square root

Known: a square plus b square equals 14, a plus B equals 4, find AB? A square b square

A²+B²=14
A+B=4
Two sides square
(A+B)²=16
A²+2AB+B²=16
14+2AB=16
∴AB=1
A²B²=(AB)²=1

Let real numbers a, B, C, satisfy a + B + C-2A ^ 2 / 1-2 (B + 1) ^ 2 / 1-2 (C-1) 2 / 1 + 3 = 0, find the value of a ^ 2 + B ^ 2 + C ^ 2 + AB + BC + AC

Original formula = (A-2 √ a + 1) + ((B + 1) - 2 √ (B + 1) + 1) + ((C-1) - 2 √ (C-1) + 1)
=(√a-1)^2+(√(b+1)-1)^2+(√(c-1)-1)^2>=0
The equal sign holds if and only if all three square terms are 0
So a = 1, B = 0, C = 2
So a ^ 2 + B ^ 2 + C ^ 2 + AB + AC + BC = 7

It is known that the real numbers a > 0, b > 0, a (a, 1), B (2, b), C (4, 5) are three points on the coordinate plane. If AC ⊥ BC, then the maximum value of AB is______ ．

∵ a (a, 1), B (2, b), C (4, 5) ∵ AC = (4-A, 4), BC = (2, 5-b) ∵ AC ⊥ BC ∵ AC · BC = 2 (4-A) + 4 (5-b) = 0 ∵ a + 2B = 14 from the basic inequality, a + 2B ≥ 22ab ∵ ab ≤ 492 can be obtained, that is, the maximum value of AB is 492, so the answer is: 492

a. If B. C is a positive real number, then the maximum value of (AB + BC) \ (a ^ 2 + B ^ 2 + C ^ 2) is?

From the proposition and "basic inequality", we can get: A & # 178; + (B & # 178 / 2) ≥ (√ 2) ABC & # 178; + (B & # 178 / / 2) ≥ (√ 2) BC equal sign only if B = (√ 2) a = (√ 2) C. by adding two inequalities, we can get: A & # 178; + B & # 178; + C & # 178; ≥ (√ 2) (AB + BC) > 0 ≠ 0 ＜ (AB + BC) / (A & # 178; + B

a. B, C are real numbers, and ab + BC + Ca = 1, find the maximum or minimum value of 1 / A + 1 / B + 1 / C. The maximum or minimum value of a + B + C
A, B, C are all positive real numbers. AB + BC + Ca = 1, use the basic inequality to find the maximum or minimum value of 1 / A + 1 / B + 1 / C. The maximum or minimum of a + B + C

The binary implicit function is derived from ab + BC + Ca = 1, which is changed into an explicit function C = (1-ab) / (a + b), which is substituted into the following two formulas
(a + b) / (1-ab) + 1 / B + 1 / C, the partial derivatives of B and C are fa = (1 + B ^ 2) / (1-ab) ^ 2-1 / A ^ 2, FB = (1 + A ^ 2) / (1-ab) ^ 2-1 / b ^ 2, at the same time, the two partial derivatives are equal to 0, a ^ 2 + 2ab-1 = 0, B ^ 2 + 2ab-1 = 0, the solution is a = b = √ 3 / 3, C = √ 3 / 3, the original formula is 3 √ 3
Substituting into the second formula, we can get (1-ab) / (a + b) + A + B, finding the partial derivative, we can get FA = 1 - (b ^ 2 + 1) / (a + b) ^ 2, FB = 1 - (a ^ 2 + 1) / (a + b) ^ 2, making the two partial derivatives equal to 0, we can get a ^ 2 + 2ab-1 = 0, B ^ 2 + 2ab-1 = 0, which is the same as the above equation, so a = b = C = √ 3 / 3, so the original formula = √ 3
Note: when the partial derivative is 0, the solution is the extremum
I do it with advanced mathematics. You can read it if you understand it, even if you don't understand it

Given real numbers a, B, C, satisfy a ^ 2 + B ^ 2 = 1, B ^ 2 + C ^ 2 = 2, a ^ 2 + C ^ 2 = 2, find the minimum value of AB + AC + BC
Given the real numbers a, B, C, satisfy a ^ 2 + B ^ 2 = 1, B ^ 2 + C ^ 2 = 2, C ^ 2 + A ^ 2 = 2, find the minimum value of AB + BC + ca

It is known that a & sup2; + B & sup2; = 1, B & sup2; + C & sup2; = 2, a & sup2; + C & sup2; = 2
Find the minimum value of AB + AC + BC
First, according to the known conditions, the values of a, B and C are solved
It is known that,
a²+b²=1 ①
b²+c²=2 ②
a²+c²=2 ③
③ - 1. It's necessary
B & sup2; = 1 / 2, that is, B = ± 1 / √ 2
Substituting the value of B & sup2; into (1), we get
A & sup2; = 1 / 2, i.e. a = ± 1 / √ 2
Substituting the value of a & sup2; into 2, we get,
C & sup2; = 3 / 2, that is, C = ± √ 3 / √ 2
a. B and C each have two values. Because the minimum value of AB + AC + BC is required, that is, each product must be negative. According to the principle of "positive, positive, negative, positive and negative", in each product, the two values must take opposite signs
ab+ac+bc
=1/2-√3/2-√3/2
=1/2-√3
(take a = b = 1 / √ 2, C = - √ 3 / √ 2 or a = b = - 1 / √ 2, C = √ 3 / √ 2)