If x = sqrt{5} + 2,then the value of frac{2x^ | vedclass

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(C) Given $x = \sqrt{5} + 2$.First,calculate $x^2 = (\sqrt{5} + 2)^2 = 5 + 4 + 4\sqrt{5} = 9 + 4\sqrt{5}$.Substitute $x$ and $x^2$ into the numerator:

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$0.625$

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(C) Given $x = \sqrt{5} + 2$.First,calculate $x^2 = (\sqrt{5} + 2)^2 = 5 + 4 + 4\sqrt{5} = 9 + 4\sqrt{5}$.Substitute $x$ and $x^2$ into the numerator: $2(9 + 4\sqrt{5}) - 3(\sqrt{5} + 2) - 2 = 18 + 8\sqrt{5} - 3\sqrt{5} - 6 - 2 = 10 + 5\sqrt{5}$.Substitute $x$ and $x^2$ into the denominator: $3(9 + 4\sqrt{5}) - 4(\sqrt{5} + 2) - 3 = 27 + 12\sqrt{5} - 4\sqrt{5} - 8 - 3 = 16 + 8\sqrt{5}$.Now,the expression is $\frac{10 + 5\sqrt{5}}{16 + 8\sqrt{5}}$.Factor out constants: $\frac{5(2 + \sqrt{5})}{8(2 + \sqrt{5})} = \frac{5}{8}$.Calculating the decimal value: $\frac{5}{8} = 0.625$.

If $x = \sqrt{5} + 2$,then the value of $\frac{2x^2 - 3x - 2}{3x^2 - 4x - 3}$ is equal to

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