Vă rog să mă ajutați ! Mulțumesc mult de tot !!
[tex]\it Efectua\c{\it t}i:\\ \\ \dfrac{3}{x+2}+\dfrac{x}{x+2}+\dfrac{6x-x^2}{x^2-4}[/tex]
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Bună!
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[tex] \frac{3}{x + 2} + \frac{x}{x + 2 } + \frac{6x - x ^{2} }{x^{2} - 4 } = \\ \frac{3 + x}{x + 2} + \frac{6x - x ^{2} }{(x - 2)(x + 2)} = \\ \frac{(3 + x)(x - 2) + 6x - {x}^{2} }{(x - 2)(x + 2)} = \\ \frac{3x - 6 + {x}^{2} - 2x + 6x - {x}^{2} }{(x - 2)(x + 2)} = \\ \purple{ \\ \frac{7x - 6}{(x - 2)(x + 2)} }[/tex]
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Sper că te-am ajutat!
[tex]\it \dfrac{3}{x+2}+\dfrac{x}{x+2}+\dfrac{6x-x^2}{x^2-4}=\dfrac{^{x-2)}3+x}{\ \ \ x+2}+\dfrac{6x-x^2}{(x-2)(x+2)}=\\ \\ \\ =\dfrac{3x+x^2-6-2x+6x-x^2}{(x-2)(x+2)}=\dfrac{7x-6}{(x-2)(x+2)}[/tex]