Question

Amplificati cu x-3 următoarele fracții algebrice:
a)
$\frac{7}{ { \times }^{2} }$
b)
$\frac{x}{x + 1}$
c)
$\frac{1 - x}{ {x}^{2} - 3x}$
d)
$\frac{x - 3}{ {x}^{2} - 3x + 9}$
va rog urgent,am nevoie azi
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Answer 1

Explicație pas cu pas:

Salutare!

$\bf \: a) \: \dfrac{7}{ {x}^{2}} = \dfrac{7 \cdot(x - 3)}{ {x}^{2}\cdot(x - 3)} = \dfrac{7x - 21}{ {x}^{3} - 3{x}^{2}}$

$\bf \: b) \: \dfrac{x}{ x + 1} = \dfrac{x \cdot(x - 3)}{(x - 3)\cdot(x + 1)} = \dfrac{ {x}^{2}- 3x}{ {x}^{2}+ x- 3x - 3} =\dfrac{ {x}^{2}- 3x}{ {x}^{2}- 2x - 3}$

$\bf \: c) \: \dfrac{1 - x}{ {x}^{2} - 3x} = \dfrac{(1 - x) \cdot(x - 3)}{(x - 3)\cdot({x}^{2} - 3x)} = \dfrac{4x-{x}^{2}-3}{{x}^{3} - 6{x}^{2} + 9x}$

$\bf \: d) \: \dfrac{x- 3}{ {x}^{2}- 3x+9} = \dfrac{(x - 3) \cdot(x - 3)}{(x- 3)\cdot({x}^{2} - 3x+9)} = \dfrac{{x}^{2} - 6x+9}{{x}^{3} - 6{x}^{2} + 18x-27}$