$\displaystyle 3:3,6,9,12,15,18,21,24,27,30,33,36,39,42$

$\displaystyle 6:6,12,18,24,30,36,42$

$\displaystyle 7:7,14,21,28,35,42$

The smallest number they all have in common in $\displaystyle 42$.

Simplifying this expression is similar to 1/2 – 1/5.  The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2 * 5 = 10.  So the problem becomes 1/2 – 1/5 = 5/10 – 2/10 = 3/10.

$\displaystyle \dpi{100} \small \frac{0}{6}=0$, which is an integer (a number with no fraction or decimal part).  All the other choices reduce to non-integers.

$\displaystyle \frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}=\frac{x^{2}}{3y^{3}z}$

First, let's simplify $\displaystyle \frac{4}{12}$. The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\displaystyle \frac{4}{12}=\frac{1}{3}$.

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $\displaystyle \frac{1}{3}x^{2}y^{-3}z^{-1}$ or $\displaystyle \frac{x^{2}}{3y^{3}z}$

24/32 = 1.33

16/12 =1.33

224/168 =1.33

4/3 = 1.33

96/72 = 1.33

160/96 = 1.67

$\displaystyle y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}=\frac{(x-\sqrt3)(x+\sqrt5)}{x-\sqrt3}=x+\sqrt5$

The root occurs where $\displaystyle y=0$. So we substitute 0 for $\displaystyle y$.

$\displaystyle y=x+\sqrt{5}$

$\displaystyle 0=x+\sqrt{5}$

This means that the root is at $\displaystyle x=-\sqrt5$.

The correct approach to solve this problem is to first write factors for the numerator and the denominator:

$\displaystyle 5: 1, 5$

$\displaystyle 125: 1, 5, 25, 125$

The highest common factor is 5. Therefore, you can divide the numerator and denominator by 5 in order to get a simplified fraction. 

Thus the numerator becomes,

$\displaystyle \frac{5}{5}=1$ and the denominator becomes $\displaystyle \frac{125}{5}=25$.

Therefore the final answer is $\displaystyle \frac{1}{25}$.

Find the common factors of the numerator and denominator.  They both share factors of 2,4, and 8.  For simplicity, factor out an 8 from both terms and simplify.

$\displaystyle \frac{32}{88}= \frac{8\times 4}{8\times 11}= \frac{4}{11}$

Remember that when you divide a fraction by a fraction, that is the same as multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator. 

In other words,

$\displaystyle \frac{\frac{a^2}{b}}{\frac{a}{b}} = \frac{a^2}{b}\cdot\frac{b}{a}=\frac{a^2b}{ab}$

Simplifying this final fraction gives us our correct answer, $\displaystyle a$.

To solve for $\displaystyle x$, simplify the fraction. In order to do this, recall that dividing by a fraction is the same as multiplying by its reciprocal. Therefore, rewrite the equation as follows.

$\displaystyle \frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1\Rightarrow \frac{x-2}{4^2}\times \frac{2}{x^2-4}=1$

Now, simplify the first fraction by calculating four squared.

$\displaystyle \frac{x-2}{16}\times \frac{2}{x^2-4}=1$

From here, factor the denominator of the second fraction.

$\displaystyle \frac{x-2}{16}\times \frac{2}{(x-2)(x+2)}=1$

Next, factor the 16.

$\displaystyle \frac{x-2}{2(8)}\times \frac{2}{(x-2)(x+2)}=1$

From here, cancel out like terms that are in both the numerator and denominator. In this particular case that includes (x-2) and 2.

$\displaystyle \frac{1}{8(x+2)}=1$

Now, distribute the eight.

$\displaystyle \frac{1}{8x+16}=1$

Next, multiply both sides by the denominator.

$\displaystyle (8x+16)\times \frac{1}{8x+16}=1\times (8x+16)$

The (8x+16) cancels out and leaves the following equation.

$\displaystyle 1=8x+16$

Now to solve for $\displaystyle x$ perform opposite operations to move all numerical values to one side of the equation leaving $\displaystyle x$ by itself on the other side of the equation.

$\displaystyle \\1=8x+16 \\-15=8x \\\\\frac{-15}{8}=x$