Given the set M = {(x, y) | 4x + y = 6}, P = {(x, y) | 3x + 2Y = 7}, then m ∩ P equals () A. （1，2）B. {1}∪{2}C. {1，2}D. {（1，2）}

Given the set M = {(x, y) | 4x + y = 6}, P = {(x, y) | 3x + 2Y = 7}, then m ∩ P equals () A. （1，2）B. {1}∪{2}C. {1，2}D. {（1，2）}

Because 4x + y = 63x + 2Y = 7, the solution is x = 1y = 2, so m ∩ P = {(x, y) | 4x + y = 6} {(x, y) | 3x + 2Y = 7} = {(1, 2)}, so D

Given the set P = {x | X & # 178; - 2x = 0} s = {x | ax + 2 = 0} and S is a subset of P, find the value of a, thank you!

The second order of P is 0 and 2
p={0,2}
If s is an empty set
Then AX = - 2 has no solution
So a = 0
When a ≠ 0
x=-2/a
In this case, s is not an empty set
He is a subset of {0,2}
Then - 2 / a = 0 or 2
=0 has no solution
So a = - 1
So a = 0, a = - 1

Set a = {x | ax ^ 2-2x + 1 = 0, X belongs to R}, then the number of subsets of a is
Is it 2 ^ n?

It's not... It's a classified discussion

It is known that set a = {x | ax ^ 2-4x + 3 = 0}, B = {x ^ 2-2x + 1 = 0}, if a and B = a, find the value of A

Solution B = {x ^ 2-2x + 1 = 0} = {1}
From a to B = a
Let B be a subset of A
So the root of equation AX ^ 2-4x + 3 = 0 at 1
That is, A-4 + 3 = 0
The solution is a = 1

It is known that in the RT triangle ABC, AC = BC, angle c = 90 degrees, D is the midpoint of side AB, and angle EDF = 90 degrees. Angle EDF rotates around point D, and its two sides intersect AC and CB (or their extension lines) at e and f respectively. When angle EDF rotates around point d until De is perpendicular to AC and E, it is easy to prove s triangle def + s triangle CEF = 1 / 2S triangle ABC,

Method 1
∵ CD is the middle line of the hypotenuse of an isosceles right triangle, ∵ ACD = 45 ° = ∠ a,
∴AD=CD,
When de ⊥ AC, AE = CE, s Δ CDE = 1 / 2S Δ ACD,
Similarly, s Δ CDF = 1 / 2S Δ BCD,
S quadrilateral CEDF = 1 / 2S Δ ABC
Method 2
∵∠EDF=∠DEC=∠ACB=90°,
The quadrilateral decf is a rectangle,
∵ D is the midpoint of the hypotenuse ab of the isosceles right triangle Δ ABC,
With CD bisection, ACB, ECD = 45 degrees,
The Δ CDE is an isosceles right triangle, CE = De,
The rectangle decf is a square
∵ CD bisects isosceles Δ ABC,
∴SΔADE=SΔCDE=SΔCDF=SΔBDF,
S Square = 1 / 2S Δ ABC

In triangle ABC, if Jiao a = 1 / 2 angle B = 1 / 6 angle c, then the shape of triangle ABC is the same

20, 40, 120 degrees, obtuse angle

Angle a is twice of angle B, and angle c is 12 degrees larger than angle a + angle B. the shape of triangle ABC can be judged
Angle c = 90 degrees, the difference between angle A and angle B is 20 degrees, calculate angle B

Because angle a = 2, angle B
And because: angle c = angle a + angle B + 12
So angle c = 3, angle B + 12
Angle a + angle B + angle c = 180 degrees
So the angle B = 28 degrees
Angle a = 56 degrees
Angle c = 96 degrees
So the triangle ABC is an obtuse triangle

In triangle ABC, angle a is twice as big as angle B, and angle c is 12 degrees bigger than angle a + angle B. what kind of triangle is this triangle?

180-3x-3x = 12; X = 26, a = 28, B = 56, C = 96, obtuse angle

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If ∠ C = 2 / 3 * π, ABC will form an arithmetic sequence with a tolerance of 2
Find C

a. B, C in turn into arithmetic sequence, d = 2
a=a ,b= a+2,c=a+4
C= 2π/3
c^2=a^2+b^2-2abcosC
(a+4)^2 = a^2+(a+2)^2 + a(a+2)
a^2+8a+16 = 3a^2+6a+4
a^2-a-6=0
(a-3)(a+2)=0
a=3
c=a+4 = 7

If the inner angles a, B and C in the triangle ABC form an arithmetic sequence and the tolerance D ＞ 0, the opposite sides of a, B and C are a, B and C respectively. If C = 2A, find the value of D

A+B+C=B-d+B+B+d=3B=180°,B=60°,A+C=120°
It can be obtained from the sine theorem
c/sinC=a/sinA
The results show that sin (120 ° - a) = 2sina
sin120°cosA-cos120°sinA=2sinA
It can be concluded that COTA = √ 3
A = 30 ° C = 90 ° so d = 30 °
It can also be calculated by cosine theorem
b^2=a^2+c^2-2accosB=a^2+4a^2-2a^2=3a^2
It can be found that a ^ 2 + B ^ 2 = C ^ 2, so ABC is RT triangle, C = 90 °, d = C-B = 30 °