Factorization a ^ n + 1 + A ^ n-1 - 2A ^ n Such as the title

Factorization a ^ n + 1 + A ^ n-1 - 2A ^ n Such as the title

a^n+1 +a^n-1 -2a^n
=a^n-1(a^2-2a+1)
=a^n-1(a-1)^2

(x²-y)²-(2y-2)²
(a²-2b)²-(1-2b)²
factorization of polynomial

In the plane rectangular coordinate system, the straight line L: y = - 4 / 3x + 8 intersects the x-axis and y-axis with a and B respectively, and rotates △ AOB clockwise 90 ° around point O to obtain △ a ` ob`
(1) An analytic formula for finding the straight line a ` B '
(2) If line a ` B 'intersects line L at point C, calculate the area of △ a ` BC

1、A（6,0）B(0,8)
So a '(0, - 6) B' (8,0)
So the analytic expression of a'B 'is y = 3 / 4x-6
2. The intersection C (168 / 25, - 24 / 25) can be obtained from y = 3 / 4x-6, y = - 4 / 3x + 8
A'B=OB+OA'=8+6=14
The area of △ a ` BC is s = 1 / 2 * 14 * 168 / 25 = 1176 / 25

12x * 9 (10-x) = 1 to solve the equation

12X*9(10-X)=1
1080x-108x^2=1
108x^2-1080x+1=0
X = {1080 ± root (1080 ^ 2-4 * 108)} / (2 * 108) = {90 ± root (90 ^ 2-3)} / 3 = 30 ± (root 8097) / 3
X1 = 30 - (root 8097) / 3
X2 = 30 + (root 8097) / 3

Original problem: vector group A1, A2, A3 are linearly related, A2, A3, A4 are linearly independent. It is proved that A4 cannot be linearly represented by A1, A2, A3
I want to know whether A1, A2, A3 and A4 are linearly correlated. If they are correlated, then whether A4 can be linearly represented by A1, A2 and A3. If not, why?

If A4 = k1a1 + k2a2 + k3a3, ① K1 ≠ 0 (otherwise A2, A3, A4 are linearly correlated) ∵ if A1, A2, A3 are not all zero, H1, H2, H3 makes 0 = h1a1 + h2a2 + h3a3, ② H1 ≠ 0 (...)

What is the square of negative m minus 2n plus the square of M minus 2n to the second power
-（m²-2n）+（m²-2n²）

-m²-2n+m²-2n²
=-2n-2n²

(2007 Guangzhou level test) the sum of the first n terms of the arithmetic sequence {an} is known as Sn, A2 = 2, S5 = 0. (1) find the general formula of the sequence {an}; (2) when n is the value, Sn gets the maximum

(1) ∵ A2 = 2, S5 = 0, ∵ a1 + D = 25a1 + 5 × 4D2 = 0, the solution is A1 = 4, d = - 2. ∵ an = 4 + (n-1) × (- 2) = 6-2n. (2) Sn = Na1 + n (n − 1) D2 = 4N − n (n − 1) = - N2 + 5N = − (n − 52) 2 + & nbsp; 254. ∵ n ∈ n *, ∵ when n = 2 or n = 3, the maximum value of Sn is 6

Function f (x) = loga (x + 1) - 1

When 1

As shown in the figure, take sides AC and BC of △ ABC as one side respectively, and make square ACDE and cbfg outside △ ABC. Point P is the midpoint of EF. Prove that the distance from point P to AB is half of ab

Then Er ∥ PQ ∥ FS, ∵ P is the midpoint of EF, ∥ q is the midpoint of RS, ∥ PQ is the median line of trapezoidal EFSR, ∥ PQ = 12 (ER + FS), ∥ AE = AC (equal side length of square), ∥ aer = ∥ cat (equal residual angle of the same angle), ∥ r = ∥ ATC = 90 °, ∥ RT △ aer ≌ RT △ cat (AAS), the same as RT △ BFS ≌ RT △ CB T，∴ER=AT，FS=BT，∴ER+FS=AT+BT=AB，∴PQ=12AB．

If the sum of the two factors of the quadratic trinomial ax + 2x-1 is 4x, then a is equal to X

Obviously, both factors are linear
So let one be MX + n
Then the other is 4x-mx-n
(mx+n)[(4-m)x-n]
=m(4-m)x^2+[(4-m)n-mn]x-n^2
=ax^2+2x-1
m(4-m)=a
[(4-m)n-mn]=2
-n^2=-1
n^2=1
If n = 1, then
(4-m)n-mn=4n-2mn=2
4-2m=2
m=1
a=m(4-m)=3
If n = - 1, then
(4-m)n-mn=4n-2mn=2
-4+2m=2
m=3
a=m(4-m)=3
So a = 3