Let a = (x belongs to R | ax2-3x-2 = 0) if a is the total set of elements, find the sum of a and a Such as the title

Let a = (x belongs to R | ax2-3x-2 = 0) if a is the total set of elements, find the sum of a and a Such as the title

When a = 0, x = - 2 / 3, then a = {- 2 / 3}
When a ≠ 0, if a is a single element set, the equation AX ^ 2-3x-2 = 0 must have two equal roots
Δ = 0, i.e. 9 + 8A = 0, a = - 9 / 8
Solve the equation - 9 / 8x ^ 2-3x-2 = 0 to get x = - 4 / 3, where a = {- 4 / 3}
The known set a = {x | ax-3x-4 = 0, X belongs to R, a belongs to R}
1. If a has two elements, find the value range of real number A. 2. If a has at most one, find the value
1. There are two elements in a, so ax ^ 2-3x-4 = 0 has two different solutions, so "deta" = B ^ 2-4ac = (- 3) ^ 2 + 16A > 0
9 + 16A > 0, a > - 9 / 16 2. A has at most one, so "deta" = B ^ 2-4ac = (- 3) ^ 2 + 16A = 0, a = - 9 / 16
The known set a = {x | ax-3x + 2 = 0}
If a is a set of single elements, find the value of a and set a
2 find the set P = {a belongs to R} a so that a contains at least one element
1. If a = 0, a = {2 / 3}
If a is not equal to 0, a = 9 / 8, a = {4 / 3}
2.p={a|a
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1. Classified discussion
(1) When a = 0, - 3x + 2 = 0, so x = 2 / 3, then set a = {2 / 3}
(2) When a is not equal to 0, then Δ = (- 3) ^ 2-2 * 4 * a = 0, so a = 9 / 8, then a = {4 / 3}
2. Classified discussion
(1) When there is one element, a = 0 or 9 / 8
(2) When there are two elements, then Δ = 9-8a > = 0, then a
Junior high school mathematics problem! Factorization! Urgent! Additional score!
（a²+4a)²+8（a²+4a)+16
（a²+4a)²+8（a²+4a)+16
= [（a²+4a)+4 ]²
= （a²+4a+4)²
= [（a+2)² ]²
=（a+2)^4
A = 160000000, B = 4000, then the square of a is divided by 2B =?
3.2*10^14
3.2 * 10 14
Five factoring problems,
Question 1: (A-1) (a-2b-1) + B & sup2;
Question 2: A & sup2; - B & sup3; - 2b-1
The third question: M-M & sup3; - Mn & sup2; + 2m & sup2; n
Question 4: X & sup2; + Y & sup2; - Z & sup2; - 2xy-12x-12y + 36
Question 5: A & sup2; + (a + 1) & sup2; + (A & sup2; + a) & sup2; + 1
1. Original formula = a2-2ab-a-a + 2B + B2 = a2-2ab + b2-2a + 2B + 1 = (a-b) 2-2 (a-b) + 1 = [(a-b) - 1] 2 = (a-b-1) 22. Original formula = a2 - (B2 + 2B + 1) = a2 - (B + 1) 2 = [a + (B + 1)] [a - (B + 1)] = (a + B + 1) (a-b
If a = 3 / 4 of B, find the square of a + 2Ab + 2B, the square of a-Ab + B
8. Before 2000
(a²-ab+b²)/(a²+2ab+2b²)
The numerator and denominator are divided by B & sup2;
a²/b²=(a/b)²=9/16
ab/b²=a/b=3/4
b²/b²=1
So the original formula = (9 / 16-3 / 4 + 1) / (9 / 16 + 2 × 3 / 4 + 2) = 1 / 5
thirty-seven point five six two five
You have to factorize it!
1、x(x-3)-2y(2y+3)
Factorization
2. It is known that a, B and C are the lengths of three sides of triangle ABC, and a ^ - C ^ - 2BC is proved
One
x(x-3)-2y(2y+3)
=x^-3x-4y^-6y=(x-2y)(x+2y)-3(x+2y)=(x+2y)(x-2y-3)
Two
This conclusion is wrong
For example, right triangle, a ^ - C ^ = B ^, as long as b > 2c, the conclusion is wrong
I think it should be a ^ - C ^ - 2BC
x(x-3)-2y(2y+3)=x^2-3x-4y^2-6y=(x+2y)(x-2y)-3(x+2y)=(x+2y)(x-2y-3)
x(x-3)-2y(2y+3)
＝x② - 3x – 4y② – 6y
＝x② - 4y② - ( 3x＋6y)
＝(x – 2y)(x+2y)-3(x+2y)
＝(x+2y)(x-2y-3)
② Represents square
x(x-3)-2y(2y+3)
=X squared-3x-4y squared-6y
=Xsquare-4ysquare-3x-6y
=(x+2y)(x-2y)-3(x+2y)
=(x+2y)(x-2y-3)
Given that the square of a-2ab = 3, the square of b-ab = 4, then the value of the square of 2b-a is?
a^2-2ab = 3 (1)
B ^ 2-AB = 4 has 2B ^ 2-2ab = 8 (2)
(2)-(1)
2b^2 - a^2 = 8 - 3 = 5
（1）（-2x²y³)².(xy) · (xy)³
(2)(2a+3b)(2a-b)
(3)5x²（x+1）（x-1）
（4）（2x+y-1）²
（5）59.8×60.2
（6）198²
(7) (2a) & B 4
(8) (- 2 / 3 a seven B five) △ 3 / 2A & # 178; B five
(9) (6 / 5 A & #179; X four - 0.9ax & #179;) △ 3 / 5AX & #179;
（10）（7x²y³-8x³y²z）÷8x²y²
（11）12⒀÷（3⑽ · 4⑾）
(12) (factorization)
1.25x²-16y²
2.（a-b）（x-y）-（b-a）（x+y）
3.a²-4ab+4b²
4.4+12（x-y）+9（x-y）²
Hope there is a process, if you feel trouble can also write a short process or not
If I can't type it to the 13th power, I will use ⒀ and other symbols
198*198
=(200-2)(200-2)
=200*200-2*2*200+2*2
=40000-800+4
=40004-800
=39204
Too many questions: you can only answer a few questions