Question

Ma puteti ajuta ????????

Answer 1

Răspuns:

$a) \: \frac{ {x}^{2} - x - 6 }{ {x}^{2} + 3x - 18} = \frac{ {x}^{2} + 2x - 3x - 6 }{ {x}^{2} + 6x - 3x - 18 } = \frac{x(x + 2) - 3(x + 2)}{x(x + 6) - 3(x + 6)} = \frac{(x + 2)(x - 3)}{(x + 6)(x - 3)} = \frac{x + 2}{x + 6}$

$b) \: \frac{ {x}^{2} - 9x + 20 }{ {x}^{2} + x - 30} = \frac{ {x}^{2} - 4x - 5x + 20}{ {x}^{2} + 6x - 5x - 30 } = \frac{x(x -4) - 5(x - 4)}{x(x + 6) - 5(x - 6)} = \frac{(x - 4 )(x - 5)}{(x + 6)(x - 5)} = \frac{x - 4}{x + 6}$

$c) \: \frac{ {x}^{2} + 8x + 15}{ {x}^{2} + 11x + 30} = \frac{ {x}^{2} + 5x + 3x + 15 }{ {x}^{2} + 6x + 5x + 30} = \frac{x(x + 5) + 3(x + 5)}{x(x + 6) + 5(x + 6)} = \frac{(x + 5)(x + 3)}{(x + 6)(x + 5)} = \frac{x + 3}{x + 6}$

$d) \: \frac{ {x}^{2} - x - 2 }{ {x}^{2} - 2x - 3} = \frac{ {x}^{2} + x - 2x - 2}{ {x}^{2} + x - 3x - 3 } = \frac{x(x + 1) - 2(x + 1)}{x(x + 1) - 3(x + 1)} = \frac{(x + 1)(x - 2)}{(x + 1)(x - 3)} = \frac{x - 2}{x - 3}$

$e) \: \frac{ {x}^{2} - 6x + 8}{ {x}^{2} + 2x - 24} = \frac{ {x}^{2} - 2x - 4x + 8}{ {x}^{2} + 6x - 4x - 24} = \frac{x(x - 2) - 4(x - 2)}{x(x + 6) - 4(x + 6)} = \frac{(x - 2)(x -4 )}{(x + 6)(x - 4)} = \frac{x - 2}{x + 6}$

$f) \: \frac{ {x}^{2} + 3x - 18 }{ {x}^{2} - x - 6} = \frac{ {x}^{2} + 6x - 3x - 18 }{ {x}^{2} + 2x - 3x - 6} = \frac{x(x + 6) - 3(x + 6)}{x(x + 2) - 3(x + 2)} = \frac{(x + 6)(x - 3)}{(x + 2)(x - 3)} = \frac{x + 6}{x + 2}$

$g)\:\frac{x^{2}-7x+12}{x^{2}-8x+15}=\frac{x^{2}-3x-4x+12}{x^{2}-3x-5x+15}=\frac{x(x-3)-4(x-3)}{x(x-3)-5(x-3)}=\frac{(x-3)(x-4)}{(x-3)(x-5)}=\frac{x-4}{x-5}$

$h)\:\frac{x^{2}+5x-32}{x^{2}+2x-24}=\frac{x^{2}+8x-4x-32}{x^{2}+6x-4x-24}=\frac{x(x+8)-4(x+8)}{x(x+6)-4(x+6)}=\frac{(x+8)(x-4)}{(x+6)(x-4)}=\frac{x+8}{x+6}$

$i)\:\frac{x^{2}+6x-16}{x^{2}-9x+14}=\frac{x^{2}+8x-2x-16}{x^{2}-2x-7x+14}=\frac{x(x+8)-2(x+8)}{x(x-2)-7(x-2)}=\frac{(x+8)(x-2 )}{(x-2)(x-7)}=\frac{x+8}{x-7}$

Stabilirea domeniului de definiție:

a) x²+3x-18 ≠ 0

b) x²+x-30 ≠ 0

c) x²+11x+30 ≠ 0

d) x²-2x-3 ≠ 0

e) x²+2x-24 ≠ 0

f) x²-x-6 ≠ 0

g) x²-8x+15 ≠ 0

h) x²+2x-24 ≠ 0

i) x²-9x+14 ≠ 0