Question

Descompuneți în factori:
1) (2x+5)^2-9
2) (4x+1)^2-36
3) (2x+3)^2-25
4) (4x-5)^2-4
5) x^16-1
6) x^64-1
7) x^256-1
8) x^1024-1
Arătați ca următoarele expresii sunt pozitive, oricare ar fi numărul real x:
1) x^2+2x+3
2) x^2+6x+11
3) x^2+12x+39
4) 4x^2+12x+13
5) 9x^2+24x+19
6) 25x^2+30x+17
7) 2x^2+12x+23
8) 2x^2-16x+35
9) 2x^2-8x+9
10) 2x^2+7x+11
11) 2x^2+5x+9
12) 2x^2+3x+7

Answer 1

1)  (2x+5+3)(2x+5-3) = (2x+ 8)(2x+2) = 4(x+4)(x+1)
2)  (4x+1+6)(4x+1-6) = (4x+7)(4x -5)
3)  (2x+3+5)(2x+3-5) = (2x+8)(2x-2) = 4(x+4)(x-1)
4)  (4x-5+2)(4x-5-2) = (4x-3)(4x-7)
5)  (x^8 + 1)(x^8  - 1) = (x^8 + 1)(x^4 + 1)(x^4 - 1) = (x^8 +1)(x^4 +1)(x² +!)(x+1)(x-1)
6)  (x^32 + 1)(x^32 - 1) = (x^32 + 1)(x^16 + 1)(x^8 +1)(x^4 +!)(x² +1)((x+1)(x-1)
7)  (x^256 - 1 ) = (x^128)² - 1 = (x^128 + 1)(x^64 + 1)(x^32 + 1)( x^16 + 1)(x^8 + 1)·
·(x^4 + 1)(x² + 1)(x+1)(x -1)
8)  x^1024 - 1 = (x^512)² - 1 = (x^512 + 1)(x^256 + 1)(x^128 + 1)(............)(x+1)(x-1)
1.  a= x² + 2x + 3 =(x+1)² + 2    (x+1)² ≥ 0 ⇒ a > 0
2.  b = x² + 6x + 11 = (x+3)² + 2    (x+3)² ≥ 0 ⇒ b > 0
3.  c = x² +12x +39 = (x+6)² + 3    (x+6)² ≥ 0 ⇒ c > 0
4.  d = 4x² + 12x + 13 = (2x + 3)² + 4    (2x+3)² ≥ 0 ⇒  d > 0
5.  e = 9x² + 24x + 19 = (3x +4)² + 3      (3x+4)² ≥ 0 ⇒ e > 0
6.  f = 25x² + 30x +17 = (5x + 3)² + 8    (5x+3)² ≥ 0  ⇒ f > 0
7.  g = 2x² + 12x + 23 = 2(x² + 6x + 9) + 5 = 2(x+3)² +5 ...... g > 0
8.  h = 2x² - 16x + 35 = 2(x-4)² + 3..................  h > 0
9.  i = 2x² -8x + 9 = 2(x² -4x + 4) + 1 = 2(x+2)² + 1 .......i > 0
10. j = 2x² +7x + 11 = 2(x² +7x/2 +49/16) + 39/8 = 2(x+7/4))² + 39/8........j > 0
11. k =2x² + 5x + 9 = 2(x² + 5x/2 + 25/16) + 47/8 = 2(x+ 5/4)² + 47/8 ....... k > 0
12.  l = 2x² + 3x + 7 = 2(x² + 3x/2 + 9/16) + 47/8 = 2(x + 3/4)² + 47/8........l > 0