Solving quadratic equation of three variables 1＝A（B＋C）,3＝A（4B＋C）,19＝3A（9B＋C）,

Solving quadratic equation of three variables 1＝A（B＋C）,3＝A（4B＋C）,19＝3A（9B＋C）,

1:AB+AC=1
2:4AB+AC=3
3:27AB+3AC=19
Formula 2-1: 3AB = 2, so AB value is 2 / 3; that is, AC value is 1 / 3, AC / AB = 1 / 2; C / b = 1 / 2
C=1
B=2
A=1/3

Solving quadratic equation of two variables 5x (x-1) = 2x (x + 2) - 4

5x(x-1)=2x(x+2)-4
5x²-5x=2x²+4x-4
3x²-9x+4=0
X1 = 2 / 2 (3 + radical 33)
X2 = 2 / 2 (3-radical 33)

Given the two focal points F1F2 of ellipse C, point P is on the ellipse, and Pf1 is perpendicular to F1F2, │ Pf1 │ = 4 / 3, │ PF2 │ = 14

|Is PF2 | equal to 14 / 3? Is it the standard equation for ellipse?
2a=|PF1|+|PF2|=6
a=3
(2c)^2=|F1F2|^2=|PF2|^2-|PF1|^2=20,c^2=5
So B ^ 2 = a ^ 2-C ^ 2 = 4
So the standard equation of ellipse is
When the focus is on the X axis, x ^ 2 / 9 + y ^ 2 / 4 = 1
When the focus is on the Y axis, y ^ 2 / 9 + x ^ 2 / 4 = 1
The centrifugal rate is e = C / a = √ 5 / 3

In the known triangle ABC, the opposite sides of angle ABC are a, B and C respectively. If the square of a - the square of B - the square of C = BC, then a =?

a²-b²-c²=bc,a²=b²+c²+bc
According to the cosine theorem: A & # 178; = B & # 178; + C & # 178; - 2bccosa
So - 2bccosa = BC, cosa = - 1 / 2
Because a is the internal angle of a triangle, a = 120 degrees

Given that the function f (x) = KX ^ 2-2x-3 is a monotone decreasing function in the interval [2,4], then the value range of K is ()

It is necessary to seek the derivative
f'(x)=2kx-2
If it is a monotone decreasing function on interval [2,4], then
When 2 ≤ x ≤ 4, 2kx-2 ≤ 0
k≤1/x,∵1/4≤1/x≤1/2,∴k≤1/4

It is proved that the sum of the diameters of the inscribed circle and circumscribed circle of a right triangle is equal to the sum of the two right angles

Let a right triangle ABC, a = 90 degrees,
The inscribed circle and the three sides are tangent to D, e and f (D is on the hypotenuse),
Center O, inscribed circle radius r,
Then BD = BF, CD = CE, AF = AE,
Connecting OE, of,
Then of ⊥ AB, OE ⊥ AC, < a = 90 degrees, OE = of,
The quadrilateral afoe is a square,
AE=AF=r,
AB+AC=AF+BF+AE+CE,
BC=BD+CD,
AB+AC-BC=AF+AE=2r,
Let R be the radius of circumcircle,
BC is a hypotenuse and a circumscribed circle with a diameter of 2R,
AB+AC=2R+2r,
That is, the sum of inscribed circle diameter and circumscribed circle diameter of right triangle is equal to the sum of two right angle sides

If u = R, a = {X - 2 ≤ x ≤ 3}, B = {x x ＜ - 1 or X ＞ 4}, then a ∩ (Cu b) is equal to?

The complement of B is - 1 ≤ x ≤ 4
The intersection of a and (the complement of B) is - 1 ≤ x ≤ 3
A ∩ (Cu B} = {X - 1 ≤ x ≤ 3}

It is known that ab = AC, ∠ a = 108 ° as shown in the figure, BD bisects ∠ ABC and intersects AC with D. It is proved that BC = AB + CD

In △ abd and △ EBD, be = BA ∠ abd ＝ ebdbd = BD (common edge), △abd ≌ △ EBD. (SAS) ≠ bed = ∠ a = 108 °, ADB = ∠ EDB. And ≌ AB = AC, ∠ a = 108 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

Find the maximum value of F (x) = x & # 178; + 2aX + 1 in the interval [- 1,2]

When A-1 / 2, the maximum is x = 2, f (2) = 5 + 4a

In the triangle ABC, ad bisects the angle BAC and intersects the angle BC at point D. if AB + BD = AC, the ratio of degree of angle B to angle c is

2:1
Extend AB to e, make be = BD, connect ed, EC, then AE = AC, and ad is angular bisector, △ AED ≌ △ ACD
∠AED=∠ACB
Also be = BD, ∠ AED = ∠ BDE,
So ∠ abd = ∠ AED + ∠ BDE = 2 ∠ AED = 2 ∠ ACB, that is ∠ B = 2 ∠ C