M ^ 2 + M-1 = 0, find the value of m ^ 3 + 2m ^ 2 + 2012

M ^ 2 + M-1 = 0, find the value of m ^ 3 + 2m ^ 2 + 2012

Because m ^ 2 + M-1 = 0
So m ^ 2 + M = 1
So m ^ 3 + 2m ^ 2 + 2012
=(m^3+m^2)+m²+2012
=m(m²+m)+m²+2012
=m*1+m²+2012
=m+m²+2012
=1+2012
=2013

It is known that a and B are real numbers, M = | 2A + B |, n = | 2a-b |, r = | 1-B |. If a + B < 0, AB < 0, | a | > | B | > 1, and 2m + N + r = 11, can a and B be determined

∵a+b

If the real numbers a and B satisfy the following conditions: the square of a + the square of B + 2A + 6B + 10 = 0, find the power of a to the power of 2010 - the power of B to the power of 3

A ^ 2 + B ^ 2-2a + 6B + 10 = 0, a ^ 2-2a + 1 + B ^ 2-6a + 9 = 0 (A-1) ^ 2 + (B + 3) ^ 2 = 0, a = 1, B = - 3, so: A ^ 2010-B ^ 3 = 1 ^ 2010 - (- 3) ^ 3 = 1 + 27 = 28

Let X1 and X2 be the two real roots of the equation x2-4x + K + 1 = 0 about X. Q: is there a real number k such that 3x1 · x2-x1 > x2 holds, please explain the reason

Two real number roots of the equation x2-4x + K + 1 = 0 of ∵ x, ∵ △ 16-4 (K + 1) ≥ 0, ∵ K ≤ 3, and 3x1 · x2-x1 ＞ X2, ∵ 3x1 · X2 - (x1 + x2) ＞ 0, while X1 + x2 = 4, x1 · x2 = K + 1, ∵ 3 × (K + 1) - 4 ＞ 0, ∵ k > 13, ∵ 13 ＜ K ≤ 3, ∵ exist real number k, making 3x1 · x2 ＞ x2 hold