Question

Simplifica fractiile:
a.)6xy pe 9x^2
b.)4x pe 6x^2y
c.)x+3 pe 5x+15
d.)x-3 pe 3x-9
e.)x^2-4x+4 pe x^2-2x
f.)x^2-4 pe x^2+2x

^-putere

Multumesc anticipat!!!

Answer 1

Sper ca te-am ajutat.Coroana?

Answer 2

$a) \frac{6x}{ {9x}^{2} } = \frac{6}{9} \times \frac{x}{ {x}^{2} } = \frac{2}{3} \times \frac{x}{ {x}^{2} } = \frac{ {2x}^{1 - 2} }{3} = \frac{ {2x}^{ - 1} }{3} = \frac{2 \times \frac{1}{x} }{3} = \frac{ \frac{2}{x} }{3} = \frac{2}{3x} \\ \\ b) \frac{4x}{ {6x}^{2y} } = \frac{4}{6} \times \frac{x}{ {x}^{2}y } = \frac{2}{3} \times \frac{x}{ {x}^{2}y } = \frac{ {2x}^{1 - 2} {y}^{ - 1} }{3} = \frac{ {2x}^{ - 1} {y}^{ - 1} }{3} = 2 \times \frac{1}{x} {y}^{ - 1} = \frac{2 \times \frac{1}{x} \times \frac{1}{y} }{3} = \frac{ \frac{2}{xy} }{3} = \frac{2}{3xy} \\ \\ c) \frac{x + 3}{5x + 15} = \frac{x + 3}{5(x + 3)} = \frac{1}{5} \\ \\ d) \frac{x - 3}{3x - 9} = \frac{x - 3}{3(x - 3)} = \frac{1}{3} \\ \\ e) \frac{ {x}^{2} - 4x + 4 }{ {x}^{2} - 2x } = \frac{ {(x - 2)}^{2} }{ {x}^{2} - 2x } = \frac{ {(x - 2)}^{2} }{x(x - 2)} = \frac{x - 2}{x} = 1 - \frac{2}{x} \\ \\ f) \frac{ {x}^{2} - 4 }{ {x}^{2} + 2x } = \frac{ {x}^{2} - {2}^{2} }{ {x}^{2} + 2x } = \frac{(x + 2)(x - 2)}{ {x}^{2} + 2x } = \frac{(x + 2)(x - 2)}{x(x + 2)} = \frac{x - 2}{x} = 1 - \frac{2}{x}$