1. If a > 3, then A-3 =? 2, if M + 3 + N-7 / 2 + (P-3) = 0, then p + 2n + 3M =? 2?

1. If a > 3, then A-3 =? 2, if M + 3 + N-7 / 2 + (P-3) = 0, then p + 2n + 3M =? 2?

1 .｜a-3｜>0
2. M + 3 = 0 and N + 7 / 2 = 0 and 2p-1 = 0, M = - 3, n = - 7 / 2, P = 3, so p + 2n + 3M = 3 + 2 * (7 / 2) + 3 * (- 3) = 1

Given that the real number a, B, C is an equal ratio sequence, and the projection of point P (1,0) on the straight line ax + by + C = 0 is Q, then what is the equation of Q? X ^ 2 + (y + 1) ^ 2 = 2
Given that the real number a, B, C is an equal ratio sequence and the projection of point P (1,0) on the straight line ax + by + C = 0 is Q, then what is the equation of Q?
x^2+(y+1)^2=2

Let the slope of Q (m, n) line ax + by + C = 0 be y = - A / bx-c / B, i.e. - A / B, so (n-0) / (m-1) = - 1 / (- A / b) = B / A, i.e., N / (m-1) = B / A. point Q (m, n) is on the line y = - A / bx-c / B, so n = - A / BM-C / b = - A / bm-b / a = - (m-1) / N * M-N / (m, n)

Given M-N = 3x2-2x + 1, n-p = 4-2x2, then P-M=______ ．

∵ M-N = 3x2-2x + 1, n-p = 4-2x2; ∵ M-P = x2-2x + 5; P-M = - x2 + 2x-5

A ^ 2 = 1. A ^ m = 2, a ^ n = 3, a ^ P = 4, find the value of a ^ m + N + P + 4,
Add: if 5 ^ 2x + 1 = 125, find the value of (X-2) ^ 2014 + X

A:
a^2=1.a^m=2,a^n=3,a^p=4,
So: A ^ 4 = (a ^ 2) ^ 2 = 1
So:
a^(m+n+p+4)
=(a^m)*(a^n)*(a^p)*(a^4)
=2*3*4*1
=24
5^2x+1=125,
5^(2x+1)=5^3
2x+1=3
2x=2
x=1
（x-2)^2014+x
=(1-2)^2014+1
=1+1
=2

The straight line (2m ^ 2 + M-3) x + (m ^ 2 + m) y = 4m-1 is parallel to the straight line 2x-3y = 5. Find the value of M
The formula of L1 / / L2 is
A1B2 - A2B1 = 0
A1C2 - A2C1 ≠ 0
The above equations are from the textbook. But I can't do it with this method
(2m ^ 2 + m - 3) (- 3) - 2 (m ^ 2 - M) = 0 / / the value is m = - 9 / 8 or M = 1
(2m ^ 2 + m - 3) (- 5) - 2 (- 4m + 1) ≠ 0 / m ≠ 13 / 10 and m ≠ - 1
According to the above calculation, then the value of M is - 9 / 8 or 1, but the answer is only - 9 / 8, so I have to round off 1, but I don't know how to round off 1, because it hasn't been calculated

The straight line (2m & sup2; + M-3) x + (M & sup2; + m) y = 4m-1 is parallel to the straight line 2x-3y = 5. Find the value of M
y=[-(2m²+m-3)/(m²+m)]x+(4m-1)/(m²+m).(1)
y=(2/3)x-5/3.(2)
The two lines are parallel and have the same slope, so - (2m & sup2; + M-3) / (M & sup2; + m) = 2 / 3
That is - 3 (2m & sup2; + M-3) = 2 (M & sup2; + m)
So 8m & sup2; + 5m-9 = 0
∴m=(-5±√313)/16.
How do you calculate M = - 9 / 8 or M = 1

It is known that the line AB is parallel to the x-axis, and the coordinates of different points a and B on the line are a (3,7-2m), B (2m, m-2), and the value of M is?

Parallel to the X axis, the ordinates are equal
7-2m=m-2
m=3

If the line passing through points a (2, - 2), B (5,0) is parallel to the line passing through points P (2m, 1), q (- 1, - M), then the value of M is

KAB=2/3
KPQ=(1+m)/(2m+1)
Because the two lines are parallel
SO 2 / 3 = (1 + m) / (2m + 1)
3(1+m)=2(2m+1)
3m+3=4m+2
4m-3m=3-2
m=1

Given that the line (2m * 2 + M-3) x + (m * 2-m) y = 4m-1 is parallel to the line 2x-3y-5 = 0, then the value of M is?

-9/8

Given that point a (m, 1) and point B (- 3, n) are symmetric with respect to the origin, find the value of M, n

Origin symmetry (x, y) = (- x, - y) ok

On the number axis, the point M representing the number (A / 2 + 2) and the point n representing the number (A / 3 + 3) are symmetrical about the origin, then the value of a is?

If two numbers are symmetric about the origin, then the sum of two numbers is equal to zero, that is, (A / 2 + 2) + (A / 3 + 3) = 0, and the solution is a = - 6