Simplify $\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{4}}{\sqrt{5}} \cdot \frac{\sqrt{6}}{\sqrt{7}}$ and rationalize the denominator of the resulting fraction.
The problem is to simplify $\frac{\sqrt{2}\cdot\sqrt{4}\cdot\sqrt{6}}{\sqrt{3}\cdot\sqrt{5}\cdot\sqrt{7}}$. Writing $\sqrt{6}$ as $\sqrt{2}\cdot\sqrt{3}$ shows that it is possible to cancel a $\sqrt{3}$ top and bottom. Also, simplify $\sqrt{4}$ to $2$. This gives $\frac{\sqrt{2}\cdot2\cdot\sqrt{2}}{\sqrt{5}\cdot\sqrt{7}} = \frac{4}{\sqrt{35}}$. Finally, to rationalize the denominator, multiply top and bottom by $\sqrt{35}$ to get $\boxed{\frac{4\sqrt{35}}{35}}$.
\frac{4\sqrt{35}}{35}
Level 4
Algebra
Problem: Simplify $\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{4}}{\sqrt{5}} \cdot \frac{\sqrt{6}}{\sqrt{7}}$ and rationalize the denominator of the resulting fraction.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first multiply the numerators and the denominators separately. The numerators are $\sqrt{2} \cdot \sqrt{4} \cdot \sqrt{6}$ and the denominators are $\sqrt{3} \cdot \sqrt{5} \cdot \sqrt{7}$. 

Calculating the numerator: 
$\sqrt{2} \cdot \sqrt{4} = \sqrt{8}$, and then $\sqrt{8} \cdot \sqrt{6} = \sqrt{48}$. 

Next, simplify $\sqrt{48}$: 
$\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$. 

Now, calculate the denominator: 
$\sqrt{3} \cdot \sqrt{5} = \sqrt{15}$, and then $\sqrt{15} \cdot \sqrt{7} = \sqrt{105}$. 

Thus, the expression simplifies to $\frac{4\sqrt{3}}{\sqrt{105}}$. 

Next, rationalize the denominator by multiplying both the numerator and the denominator by $\sqrt{105}$: 
The numerator becomes $4\sqrt{3} \cdot \sqrt{105} = 4\sqrt{315}$. 

Now simplify $\sqrt{315}$: 
$\sqrt{315} = \sqrt{9 \cdot 35} = \sqrt{9} \cdot \sqrt{35} = 3\sqrt{35}$. 

Thus, the numerator is $4 \cdot 3\sqrt{35} = 12\sqrt{35}$. 

The denominator becomes $\sqrt{105} \cdot \sqrt{105} = 105$. 

Now the expression is $\frac{12\sqrt{35}}{105}$. 

To simplify further, divide both the numerator and the denominator by 3: 
The numerator becomes $\frac{12}{3} \sqrt{35} = 4\sqrt{35}$, and the denominator becomes $\frac{105}{3} = 35$. 

Thus, the final answer is $\frac{4\sqrt{35}}{35}$.<|END_OF_LATENT|> The final answer is $\frac{4\sqrt{35}}{35}$.