A = {x ∈ R | X & # 178; - 3x + 4 = 0} B = {x ∈ R | (x + 1) (X & | 178; + 3x-4) = 0} A is the proper subset of C and C is the subset of B to find C

A = {x ∈ R | X & # 178; - 3x + 4 = 0} B = {x ∈ R | (x + 1) (X & | 178; + 3x-4) = 0} A is the proper subset of C and C is the subset of B to find C

A = {x ∈ R | X & # 178; - 3x + 4 = 0} = empty set
B={x∈R|(x＋1)(x²＋3x－4)＝0}={-4,-1,1}
A is the proper subset of C and C is the subset of B
So C = {- 4}, {- 1}, {1}, {- 4, - 1}, {- 4,1}, {- 1,1}, {- 4, - 1,1} [a total of 7]
If you don't understand, please hi me, I wish you a happy study!
Let a = {x (x-1) & # 178;
A={X|x^2-5x+5
(x-1)²
Calculate the square of a - the square of B / A, the square of B / a - AB / 2A 2B
(A's Square - B's Square) / (a + b) / (A's Square - AB) / (2a + 2b)
=(a+b)(a-b)/(a+b) × 2(a+b)/a(a-b)
=(2a+2b)/a
Factorization of quadratic equation of one variable
The square of X + root sign 2X-4 = 0 the second question the square of X + 2 (P-Q) - 4pq = 0
1. It is not suitable to use factorization method, but should use collocation method or formula method
2、x²+2px-2qx-4pq=0
x（x+p）-2q（x+p）=0
（x+p）（x-2q）=0
x+p=0,x-2q=0
So X1 = - P, X2 = 2q
Calculation: square of 4 (a-2b) - (2a + b) (- B + 2a)
4（a-2b）²-（2a+b）（-b+2a）
=4（a-2b）²+（2a+b）（2a-b）
=4a²-16ab+16b²+4a²-b²
=8a²-16ab+15b²
2 (the square of x-x) + half.
Can we still decompose factors? Ask: you are very smart. Thank you, although you have chosen a satisfactory answer
On factorization of quadratic equation of one variable
I've never been able to do this
X²-X-90=0
How to change to (x + 9) (X-10) = 0
. if more similar
This is cross multiplication,
19
1 －10
The product of the first column is the coefficient of the square term
The product of the second column is the value of the constant term, and then the cross multiplication and addition should be equal to the coefficient of the first term
When the square of a - the square of B = 5, ab = 2, the square of 2A - the square of 2Ab - the square of 2B
The square of 2a-the square of 2ab-2b
=The square of 2a-the square of 2b-2ab
=2 (square of A-Square of B) - 2Ab
=2*5-2*2
=10-4
=6
Factorization of quadratic equation of one variable
(1)——x^2-2x-99=0 (2)——2y(y-4)=4-y (3)——(x-5)^2+2x(x-5)=0 (4)4(x-√2）^2=3(√2-x)
(1) x^2-2x-99=0 x^2-2x+1-100=0(x-1)^2-100=0(x-1)^2=100x1=11 x2=-9(2) 2y(y-4)=4-y2y^2-7y-4=0(x-4)(2x+1)=0x1=4 x2=-0.5(3) (x-5)^2+2x(x-5)=0(x-5)(x-5+2x)=0(x-5)(3x-5)=0x1=5 x2=5/3(4) 4(x-√2）^2=3(√2-x)...
(1) x^2-2x-99=0
x^2-2x+1-100=0
(x-1)^2-100=0
(x-1)^2=100
x1=11 x2=-9
(2) 2y(y-4)=4-y
2y^2-7y-4=0
(x-4)(2x+1)=0
... unfold
(1) x^2-2x-99=0
x^2-2x+1-100=0
(x-1)^2-100=0
(x-1)^2=100
x1=11 x2=-9
(2) 2y(y-4)=4-y
2y^2-7y-4=0
(x-4)(2x+1)=0
x1=4 x2=-0.5
(3) (x-5)^2+2x(x-5)=0
(x-5)(x-5+2x)=0
(x-5)(3x-5)=0
x1=5 x2=5/3
(4) 4(x-√2）^2=3(√2-x)
4(x-√2）^2+3(x-√2)=0
(x-√2)(4x-4√2+3)=0
x1=√2 x2=√2-3/4
If you don't mind, you can check it_ ∩ o ∩ put away
First simplify and then evaluate (2a-3b) (- 2b-3b) + (- 2A + B's Square), where a = 1 / 2, B = 1
-Simplify yourself
Quadratic equation of one variable (factorization method)
2x+6=(x+3)²
2(x+3)=(x+3)²
How does (x + 3) & sup2; move to the right?
........]
Wrong... Move to the left... Move to the left of the equation
2x+6=(x+3)²
2x+6-(x+3)^2=0
2(x+3)-(x+3)^2=0
(x+3)(2-x-3)=0
X = - 3 or x = - 1