How to find a complex fraction

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PSAT Math › How to find a complex fraction

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1

Simplify:

$\frac{1- \frac{2}{x}}{1+ \frac{2}{x^{2}}}$

$\frac{ x^{2}-2x}{ x^{2}+2}$

CORRECT

$\frac{x}{x+2}$

0

$\frac{x-2}{x+2}$

0

$\frac{ x^{2}-2 }{ x^{2}+2}$

0

$\frac{ x^{2}+2x}{ x^{2}+2}$

0

Explanation

Simplify the numerator and the denominator, then divide, as follows:

$\frac{1- \frac{2}{x}}{1+ \frac{2}{x^{2}}}$

$=\frac{\frac{1 \cdot x}{x} - \frac{2}{x}}{\frac{1 \cdot x^{2}}{x^{2}} + \frac{2}{x^{2}}}$

$=\frac{\frac{ x}{x} - \frac{2}{x}}{\frac{x^{2}}{x^{2}} + \frac{2}{x^{2}}}$

$=\frac{\frac{ x-2}{x} }{\frac{x^{2}+2}{x^{2}} }$

$= \frac{ x-2}{x} \div \frac{x^{2}+2}{x^{2}}$

$= \frac{ x-2}{x} \cdot \frac{x^{2}}{x^{2}+2}$

$= \frac{ x-2}{1} \cdot \frac{x }{x^{2}+2}$

$= \frac{ x^{2}-2x}{ x^{2}+2}$

2

Evaluate:

$\frac{3+\frac{1}{4}}{3-\frac{1}{4}}$

$\frac{13}{11}$

CORRECT

$\frac{11}{13}$

0

$\frac{132}{16}$

0

$\frac{16}{132}$

0

0

Explanation

Simplify the numerator and the denominator, then divide, as follows:

$\frac{3+\frac{1}{4}}{3-\frac{1}{4}}$

$=\frac{ \frac{12}{4}+\frac{1}{4}}{\frac{12}{4}-\frac{1}{4}}$

$=\frac{ \frac{13}{4}}{\frac{11}{4} }$

$= \frac{13}{4}\div \frac{11}{4}$

$= \frac{13}{4}\cdot \frac{4}{11}$

$= \frac{13}{1}\cdot \frac{1}{11}$

$= \frac{13}{11}$

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