It is known that a (1, - 2) B (2,1) C (3,2) d (- 2,3) tries to express vector AD + BD + CD based on vector AB and AC With analysis, thanks

It is known that a (1, - 2) B (2,1) C (3,2) d (- 2,3) tries to express vector AD + BD + CD based on vector AB and AC With analysis, thanks

AB＝（1,3）,AC＝（2,4）.AD+BD+CD＝（-12,8）
（-12,8）＝x（（1,3）+y（2,4）.
x+2y＝-12,3x+4y＝8,x＝-16/5,y＝22/5
AD+BD+CD＝（-16/5）AB+（22/5）AC

As shown in the figure, D is a point on the extension line of BC, the bisector of angle ABC and angle ACD intersects at e, and the angle a = 58 ° is obtained

∠E=∠ECD-∠EBD=0.5∠ACD-0.5∠ABC=0.5*（∠ACD-∠ABC）=0.5*∠A=29°

Let F1F2 be the two focal points of the ellipse 4x / 49 + Y / 6 = 1, Q be the point on the ellipse, and the Mo ratio of Pf1 to PF2 is 4:3, then the area of the triangle pf1f2 is smaller

What is p?
Suppose P is Q
x²/(49/4)²+y²=6=1
a²=49/4,b²=6
c²=25/4
So a = 7 / 2, C = 5 / 2
Definition of ellipse
PF1+PF2=2a=7
PF1:PF2=4:3
So Pf1 = 4, PF2 = 3
F1F2=2c=5
It's a right triangle, obviously
So area = 6

When a triangle is not a right triangle, what is the formula for finding the radius of inscribed circle and circumscribed circle?
Write the process if you can
You can't forget it
To junior high school level to understand

Junior high school can only solve right triangle. Non right angle, need to use high school triangle knowledge, Helen formula and so on

Given the function f (x) = e ^ x-kx ^ 2, X ∈ R
(1) If the function y = f (x) increases monotonically on (negative infinity, positive infinity), find the value range of real number K
(2) If for any t ∈ (0,1], the equation f (x) = t has three different real number solutions, the value range of real number k is obtained

Given the function f (x) = e ^ x-kx ^ 2, X ∈ R
(1) If the function y = f (x) increases monotonically on (negative infinity, positive infinity), find the value range of real number K
(2) If for any t ∈ (0,1], the equation f (x) = t has three different real number solutions, the value range of real number k is obtained
(1) Analysis: ∵ function f (x) = e ^ x-kx ^ 2
Let f '(x) = e ^ x-2kx > = 0 = = > k = k' (x) = e ^ x (2x-2) / (2x) ^ 2 = 0 = = > x = 1
When 0k > e / 2, f '(x) has two zeros when x = ln (2k) takes the minimum value E / 2
That is, the function f (x) has a maximum on (0, ln (2k)) and a minimum on (LN (2k), + ∞)
The function f (x) and the function y = t (t ∈ (0,1]) must have three intersections, that is, the equation f (x) = t always has three different real solutions
The value range of real number k is k > e / 2

Verification: the sum of the two right angles of a right triangle is equal to the sum of the diameter of its circumscribed circle and the diameter of its inscribed circle

Let the two right sides of a right triangle be a and B, and the hypotenuse be c
C & # 178; = A & # 178; + B & # 178; circumscribed circle diameter d = C, inscribed circle diameter d = 2Ab / (a + B + C)
∴(a＋b＋c)(a＋b－c)＝(a＋b)²－c²＝a²＋b²＋2ab－c²＝2ab
∴a＋b－c＝2ab／(a＋b＋c)
∴a＋b＝c＋2ab／(a＋b＋c)
That is, the sum of the two right angles of a right triangle is equal to the sum of the diameter of its circumscribed circle and the diameter of its inscribed circle

Given the complete set u = = {A-1}, (A-2) (A-1), 4,6}, (1) if Cu (cub) = {0,1}, find the value of real number a (2) if CUA = {3,4}, find the value of real number a

(1) So {A-1 |, (A-2) (A-1)} = {0,1}, because A-1 is not 0 (once it is zero, there are multiple terms), so A-2 = 0, a = 2 (2) only | A-1 |, or (A-2) (A-1) = 3, if | A-1 | = 3, a = 4 or - 2, then (A-2) (A-1) = 6 or 12, (A-2) (A-1) = 6, there are multiple terms, so a = - 2, if (A-2) (A-1) = 3, two roots are not

In the space quadrilateral ABCD, ab = BC = CD = Da = AC = BD, then the cosine value of AD and plane ABC is

Take the midpoint e of BC
With AE vertical BC
BC vertical de
AED with BC vertical plane
Face ABC vertical AED
So the angle between AD and plane ABC is DAE
cos DAE= (AE^2+AD^2-DE^2)/(2AE*DA)
Known AE = EC cos DAE = ad / 2ae = radical 3 / 3

If the inequality X & # 178; - 2aX + a > 0 is r-constant for X, find the inequality a ^ 2T + 1 about t

The inequality X & # 178; - 2aX + a > 0 is r-constant for X
Then Δ = 4A & # 178; - 4A

As shown in the figure, ab = ad, AC = AE, ∠ DAB = ∠ CAE, be and DC intersect at point P

It is proved that: the crossing point a is am ⊥ DP, the perpendicular foot is m, an ⊥ PE, the perpendicular foot is n, ∵ DAB = ∠ CAE (known), and ∵ DAB + ∠ BAC = ∠ CAE + ∠ BAC (properties of the equation), that is, ∵ DAC = ∠ BAE