A = 4mn-n2, B = - 3m2n2, C = 1-4n2, where M = 2, n = 3, find the value of a + B + C

A = 4mn-n2, B = - 3m2n2, C = 1-4n2, where M = 2, n = 3, find the value of a + B + C

A
=4×2×3-3²
=15
B
=-3×2²×3²
=-108
C
=1-4×3²
=-35
A+B+C
=15-108-35
=-128

If M and N are integers, please prove that N2 + N2 (n + 1) 2 + (n + 1) 2 = (N2 + N + 1) 2

Is n ^ 2 + n ^ 2 (n + 1) ^ 2 + (n + 1) ^ 2 = (n ^ 2 + N + 1) ^ 2? Expand both sides, the left side is n ^ 2 + n ^ 2 (n ^ 2 + 2n + 1) + (n ^ 2 + 2n + 1) = n ^ 4 + 2n ^ 3 + 3N ^ 2 + 2n + 1, the right side is n ^ 4 + 2n ^ 3 + 3N ^ 2 + 2n + 1

The square of parabola y = x + MX + M-3 intersects with X axis at two points a and B, intersects with y axis at point C, and the vertex is m. (1) judge which parabola point m is on (not affected by M) (2) find the highest or lowest position of point m (3) under the condition (2), prove: Triangle OAC ∽ triangle OCB

(1) Point m is on the parabola with the opening upward
(2) The Y coordinate of point m is - m ^ / 4 + M-3 (^ means square)
This coordinate has a maximum = - B ^ / 4A + C = 1 ^ / 4 (1 / 4) - 3 = - 2
In this case, M = - B / 2A = 1 / 2 (1 / 4) = 2, the parabolic equation is y = x ^ + 2x-1 = (x + 1) ^ - 2
The highest position of point m is (- 1, - 2)
(3) To solve the parabolic equation: x = - 1 ± √ 2; the coordinates of point C are (0, - 1)
∵∠ AOC = ∠ cob = 90 degrees
AO/CO=(-1-√2)/-1=√2+1
CO/BO=1/(√2-1)=√2+1 ∴ΔOAC∽ΔOCB

It is known that the parabola y = the square of MX - 2mx + m + 2. When m is the value, the parabola is above the x-axis

y=m(x^2-2x+1)+2
y=m(x-1)^2+2
So m > 0

The square + (m-1) x + m of the parabola y = - x intersects the Y axis at the point (0,3),
1. Find the value of M and the analytical formula of parabola
2. Find the intersection of parabola and X axis and the coordinates of the intersection

1. Because it intersects with the Y axis at the point (0,3), when x = 0, y = 3, so the analytic formula of M = 3 is y = - x ^ 2 + 2x + 3
2. - x ^ 2 + 2x + 3 = 0, the solution is x = - 1 or 3
So the intersection coordinates are (- 1,0) and (3,0)

Given the parabola y = x square + MX + m-5, when m is the value, the two intersections a (x1,0), B (x2,0) of the parabola and X-axis

That is, square B - 4ac > 0,
Then: M2 - 4m + 20 > 0
When m is any real number, the expression is greater than 0
Therefore, when m is any real number, the two intersections a (x1,0), B (x2,0) of parabola and X-axis

If x2 + MX-15 = (x + 3) (x + n), then M=______ ，n=______ ．

∵ (x + 3) (x + n) = x2 + (3 + n) x + 3N, ∵ x2 + MX-15 = x2 + (3 + n) x + 3N, ∵ M = 3 + N, 3N = - 15, the solution is m = - 2, n = - 5. So the answer is - 2, - 5

The square of M minus 4Mn plus 12n minus 9 is factorized

m^2-4mn+12n-9
=(m^2-9)-(4mn-12n)
=(m+3)(m-3)-4n(m-3)
=(m-3)(m+3-4n)

(m-2n) & sup2; - 2 (2n-m) (M + n) + (M + n) & sup2; factorization

(m-2n)²-2(2n-m)(m+n)+(m+n)²
=(m-2n)²+2(m-2n)(m+n)+(m+n)²
=[(m-2n)+(m+n)]^2
=(2m-n)^2

(M & sup2; - 1) x & sup2; + (M & sup2; + 4m + 5) x + 5m factorization

m+1 m
m-1 5