Respuesta :

[tex] \bf 7\sqrt{3}-4\sqrt{6}+\sqrt{48}-\sqrt{54}~~
\begin{cases}
48=2\cdot 2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 2^2\cdot 3\\
\qquad (2^2)^2\cdot 3\\
54=2\cdot 3\cdot 3\cdot 3\\
\qquad 2\cdot 3\cdot 3^2\\
\qquad 6\cdot 3^2
\end{cases}
\\\\\\
7\sqrt{3}-4\sqrt{6}+\sqrt{(2^2)^2\cdot 3}-\sqrt{6\cdot 3^2}\implies 7\sqrt{3}-4\sqrt{6}+2^2\sqrt{3}-3\sqrt{6}
\\\\\\
7\sqrt{3}-4\sqrt{6}+4\sqrt{3}-3\sqrt{6}\implies \stackrel{\textit{adding like terms}}{11\sqrt{3}-7\sqrt{6}} [/tex]

Numenius

To simplify the above expression, let's simplify each radicals at first.

√3 and √6 cannot be further simplify.

Now √48 = √(16* 3)

=√16 * √3

= 4√3

Similarly, √54 = √(9*6)

=√9 *√6

=3√6

So, now we can write the above expression as:

7√3 - 4√6 + √48 - √54

= 7√3 - 4√6 + 4√3 - 3√6

= 11√3 - 7√6

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