Question

If x = 3-root5 the find (Root x)/(root 2 + root 3x-2)
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

1/√5

Step-by-step explanation:

given:- x = 3-√5

we have to find $\sf{\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}}}$

=> √x = √(3-√5)

=> √x = $\sf{\sqrt{\frac{2(3-\sqrt{5})}{2}}}$

=> √x = $\sf{\sqrt{\frac{6-2\sqrt{5}}{2}}}$

=> √x = $\sf{\sqrt{\frac{1^2 + \sqrt{5}^2 - 2\sqrt{5}}{2}}}$

=> √x = $\sf{\frac{\sqrt{5}-1}{\sqrt{2}}}$

=> √x = $\bold{\frac{(\sqrt{5}-1)}{\sqrt{2}}}$

also, √(3x-2) = √{3(3-√5)-2}

=> √(3x-2) = √(9-3√5-2)

=> √(3x-2) = √{2(7-3√5)/2}

=> √(3x-2) = √{(14-6√5)/2}

=> √(3x-2) = √{3² + (√5)² - 2*3√5}/2

=> √(3x-2) = √{ (3-√5)²/2}

=> √(3x-2) = (3-√5)/√2

so,

$\sf{\frac{\sqrt{x}}{\sqrt{2} - \sqrt{3x - 2}}}$

=> $\sf{\frac{\frac{\sqrt{5}-1}{\sqrt{2}}}{\sqrt{2}+\frac{3-\sqrt{5}}{\sqrt{2}}}}$

=> $\sf{\frac{\sqrt{5}-1}{2+(3-\sqrt{5})}}$

=> $\sf{\frac{\sqrt{5}-1}{5-\sqrt{5}}}$

=> $\sf{\frac{\sqrt{5}-1}{5-\sqrt{5}} *\frac{5+\sqrt{5}}{5+\sqrt{5}}}$

=> $\sf{\frac{5\sqrt{5}-5 +5 -\sqrt{5}+5}{25-5}}$

=> $\sf{\frac{4\sqrt{5}}{20}}$

=> $\bold{\sqrt{5}/5}$

=> $\bold{1/\sqrt{5}}$

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