from: category_eng

1.

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Simplify $sqrt[3]{xsqrt[3]{xsqrt[3]{xsqrt{x}}}}$ .

$mathrm{(A) } sqrt{x}qquad mathrm{(B) } sqrt[3]{x^{2}}qquad mathrm{(C) } sqrt[27]{x^{2}}qquad mathrm{(D) } ...$

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2.

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Which of the following is equivalent to $sqrt{frac{x}{1-frac{x-1}{x}}}$ when $x < 0$ ?

$mathrm{(A) } -xqquad mathrm{(B) } xqquad mathrm{(C) } 1qquad mathrm{(D) } sqrt{frac{x}{2}}qquad mathrm{(E)...$

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3.

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Assume that $x$ is a positive real number. Which is equivalent to $sqrt[3]{xsqrt{x}}$ ?

$mathrm{(A)} x^{1/6}qquadmathrm{(B)} x^{1/4}qquadmathrm{(C)} x^{3/8}qquadmathrm{(D)} x^{1/2}qquadmathrm{(E)} x$

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4.

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What is the product of all the roots of the equation $sqrt{5 | x | + 8} = sqrt{x^2 - 16}.$

$extbf{(A)} -64 qquad extbf{(B)} -24 qquad extbf{(C)} -9 qquad extbf{(D)} 24 qquad extbf{(E)} 576$

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1.

$sqrt[3]{xsqrt{x}}=sqrt[3]{(sqrt{x})^{2}cdotsqrt{x}}=sqrt[3]{(sqrt{x})^{3}}=sqrt{x}$ .

Therefore:

$sqrt[3]{xsqrt[3]{xsqrt[3]{xsqrt{x}}}}=sqrt[3]{xsqrt[3]{xsqrt{x}}}=sqrt[3]{xsqrt{x}}= sqrt{x} Rightarrow A$

2.

$sqrt{frac{x}{1-frac{x-1}{x}}} = sqrt{frac{x}{frac{x}{x}-frac{x-1}{x}}} = sqrt{frac{x}{frac{x-(x-1)}{x}}} = sqrt{...$

Since $x<0$

$|x|= -x Rightarrow A$

3.

$sqrt[3]{xsqrt{x}}=sqrt[3]{sqrt{x^3}}=sqrt[6]{x^3}=x^{3/6}=x^{1/2} mathrm{(D)}$

4.

First, simplify the fractions.

$dfrac{2+4+6}{1+3+5} - dfrac{1+3+5}{2+4+6} = dfrac{12}{9} - dfrac{9}{12}$

$dfrac{12}{9} - dfrac{9}{12} = dfrac{48}{36} - dfrac{27}{36} = dfrac{21}{36} = oxed{dfrac{7}{12} extbf{(C)}}$