If the image of function y = x square + Mn-N passes through point P (1, 2), then N-M is

If the image of function y = x square + Mn-N passes through point P (1, 2), then N-M is

Wrong. It's MX
x=1,y=2
Then 2 = 1 + M-N
So N-M = - 1
Given the complete set I = R, if the function f (x) + x2-3x + 2, the set M = f (x) is less than or equal to 0, n = sub-F (x) is less than 0, find the intersection of the complement of M and N in I
Complements of M and N union sets
Your question is somewhat ambiguous. The formula is not right. There is no equal sign in F (x). Let's consider for a moment: F (x) = x ^ 2-3x + 2. M = (1,2), speechless, what's the second meaning in n? I don't want to answer. It's not difficult. Just draw a Wayne map.
Extraction of common factor factorization: a (a-b) + (a-b) & #178;
a（a-b）+（a-b）²
=(a-b)(a+a-b)
=(a-b)(2a-b)
a（a-b）+（a-b）²
=（a-b)[a+(a-b)]
=(a-b)(2a-b)
Decomposition factor: 4a-1 = b2-4a2 (2 is square)
1. (1) 3x + 3Y (2) - 24m & sup2; x-16n & sup2; X (3) x & sup2; - 1 (4) (XY) & sup2; - 1 (5) a to the fourth power X & sup2; - A to the fourth power Y & sup2;
(6)3x²+6xy+3y² (7)(x-y)²+4xy (8)4a²-3b(4a-3b) (9)2x(a-2)-y(2-a)
(10) The fourth power of X (x + y) - Y (x + y) (11) (a + b) & sup2; + 2 (a + b) + 1 (13) 4x - 4x & sup3; + X & sup2; (14) x & sup2; - 16ax + 64a & sup2;
(15)(x-1)(x-3)+1 (16)(ab+a)+(b+1)
(1)3（x+y） (2)-8x(3m²+2n²) (3)(x+1)(x-1) (4)(xy+1)(xy)-1
It's troublesome to answer math questions
Overall, it's not difficult
Let's cheer for the second year of junior high school
A times the square of x-3bx-5 = 0 (a is not equal to 0), then what is the value of 4a-6b?
When x = 2, 2 ^ 2a-3 * 2 * B-5 = 0
That is, 4a-6b = 5
Integral multiplication and division and factorization
1. It is known that the trilateral lengths a, B and C of △ ABC satisfy the relation - C & sup2; + A & sup2; + 2ab-2bc = 0, and try to explain that △ ABC is an isosceles triangle
2.1 × 2 × 3 × 4 + 1 = 25 = 5 & sup2;; 2 × 3 × 4 × 5 + 1 = 121 = 11 & sup2;; 3 × 4 × 5 × 6 + 1 = 361 = 19 & sup2;... According to the above rule, Xiaoqiang conjectures that the sum of the product of any four continuous positive integers and one must be a complete square. Is Xiaoqiang's conclusion correct? If it is correct, please prove it; if not, please explain the reason
Because - C & sup2; + A & sup2; + 2ab-2bc = 0
So B & sup2; + A & sup2; + 2Ab = B & sup2; + 2BC + C & sup2;
So (a + b) & sup2; = (B + C) & sup2;
And a + b > 0, B + C > 0
So a + B = B + C
So a = C
So it's an isosceles triangle
Let the first of four consecutive numbers be n, then
n(n+1)(n+2)(n+3)+1
=n(n+3)(n+1)(n+2)+1
=(n² +3n)(n² +3n+2)+1
=(n² +3n)²+2(n²+3n)+1
=(n²+3n+1)²
Therefore, the sum of the product of any four consecutive positive integers and one must be a complete square number
The original solution = (A-C) (a + C) + 2B (A-C) = 0 = (A-C) (a + C + 2b) from the meaning of the title, a-c = 0, so a = C, so it is an isosceles triangle
4a2 + B2 + 4a-6b + 10 = 0 find the value of 2A2 + 4b-3
4a²+4a+b²-6b+10=0
(4a²+4a+1)+(b²-6b+9)=0
(2a+1)²+(b-3)²=0
a=-1/2,b=3
2a²+4b-3
=2×(-1/2)²+4×3-3
=1/2+9
=19/2
On the practice of "integral multiplication and division and factorization" in mathematics of grade two
Proof: for any number x and y, the value of integral X & sup2; + 2x + Y & sup2; - 6y + 12 must be positive
The above formula can be reduced to (X & sup2; + 2x + 1) + (Y & sup2; - 6y + 9) + 2
R and then after the disease, you will find that it is equal to (x + 1) & sup2; + (Y-3) & sup2; + 2
（x+1)²>=0
(y-3)²>=0
Therefore, the above formula is always positive and greater than or equal to 2
Given 4a2 + b2-4a-6b + 10 = 0, find (2 / 3a √ 9A + B2 √ A / B3) - (A2 √ 1 / a-5a √ B / a)
That is, (4a & # 178; - 4A + 1) + (B & # 178; - 6B + 9) = 0
(2a-1)²+(b-3)²=0
So 2a-1 = B-3 = 0
a=1/2,b=3
So the original formula is 2A √ a + √ AB-A √ a-5 √ ab
=a√a-4√ab
=√2/4-4√6/2
=√2/4-2√6