Răspuns:
Explicație pas cu pas:
b) (4x+8)/(x²+4x+4) × ( x²-4)/(4x-8) = 1
= 4(x+2)/(x+2)² × [(x-2)(x+2)]/4(x-2) =
= [(x+2)(x+2)]/(x+2)² × 4(x-2)/4(x-2) =
= 1
__________________________________________________
c)
(x²+4x)/(x²-16) × (3x-12)/(5x+25) =
= x(x+4)/[(x+4)(x-4)] × 3(x-4)/(5x+25) =
= [(x+4)(x-4)]/[(x+4)(x-4)] × 3 x/(5x+25) =
= 3x/5(x+5) x ≠ - 5
______________________________________
e)
= 7(x+5)/[(x+5)(x-5)] × [(x-5)(x-2)]/(x-2)² =
= 7/(x-2) unde x≠2
____________________________________
f)
= 3x(x-4)/(x-4)² × [(x-4)(x-3)]/(x-3)² =
(x-4)(x-4) /(x-4)² = 1
(x-3)/(x-3)² = 1/(x-3)
= 3x/(x-3); x ≠ 3
_____________
3x²- 12 x = 3x ( x - 4) -> l-am dat factor comun pe 3 x
x² - 8 x + 16 = (x-4)²
x² - 7 x + 12 = x² - 4 x - 3 x + 12 = x(x-4) - 3(x-4) = (x-4)(x-3)
x² - 6 x + 9 = ( x-3)²
