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[tex]\frac{1}{x {}^{2} - 3 } = \frac{ \sqrt{3} }{x + \sqrt{3} } - \frac{2}{ \sqrt{3} - x } [/tex]
can someone please solve this?

Respuesta :

Answer:

[tex]\boxed{\textsf{ The value of x is \textbf{ 8 + 6}$\sqrt{3 }$.}}[/tex]

Step-by-step explanation:

Here we are given a equation which we need today simplify and solve for x .The given equation to us is :-

[tex]\sf\implies \dfrac{1}{x {}^{2} - 3 } = \dfrac{ \sqrt{3} }{x + \sqrt{3} } - \dfrac{2}{ \sqrt{3} - x } [/tex]

Let's simplify the equation :-

[tex]\sf\implies \dfrac{1}{x {}^{2} - 3 } = \dfrac{ \sqrt{3} }{x + \sqrt{3} } - \dfrac{2}{ \sqrt{3} - x } \\\\\sf\implies \dfrac{1}{x^2-3}=\dfrac{\sqrt3}{x+\sqrt3}-\dfrac{2}{-(x-\sqrt3)}\\\\\sf\implies \dfrac{1}{x^2-3}=\dfrac{\sqrt3}{x+\sqrt3}+\dfrac{2}{x-\sqrt3}\\\\\sf\implies \dfrac{1}{x^2-3}=\dfrac{\sqrt3(x-\sqrt3)+2(x+\sqrt3)}{(x+\sqrt3)(x-\sqrt3)}\\\\\sf\implies \dfrac{1}{x^2-3}=\dfrac{\sqrt3x-3+2x+2\sqrt3}{x^2-3}\\\\\sf\implies 1=\sqrt3x-3+2x+2\sqrt3 \\\\\sf\implies \sqrt3x+2x = 4-2\sqrt3\\\\\sf\implies x(\sqrt3 +2)=4-2\sqrt3\\\\\sf\implies x =\dfrac{4-2\sqrt3}{\sqrt3+2}\\\\\sf\implies x =\dfrac{ (4-2\sqrt3)(\sqrt3-2)}{(\sqrt3+2)(\sqrt3-2)}\\\\\sf\implies x =\dfrac{-8-6\sqrt{3}}{3-4}\\\\\sf\implies \boxed{\pink{\sf x =8+6\sqrt3}}[/tex]

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