Respuesta :

Answer:

5[tex]\sqrt{3}[/tex]

Step-by-step explanation:

Using the rule of radicals

[tex]\sqrt{a}[/tex] × [tex]\sqrt{b}[/tex] ⇔ [tex]\sqrt{ab}[/tex]

Simplifying the radicals

[tex]\sqrt{27}[/tex] = [tex]\sqrt{9(3)}[/tex] = [tex]\sqrt{9}[/tex] × [tex]\sqrt{3}[/tex] = 3[tex]\sqrt{3}[/tex]

[tex]\sqrt{48}[/tex] = [tex]\sqrt{16(3)}[/tex] = [tex]\sqrt{16}[/tex] × [tex]\sqrt{3}[/tex] = 4[tex]\sqrt{3}[/tex]

[tex]\sqrt{12}[/tex] = [tex]\sqrt{4(3)}[/tex] = [tex]\sqrt{4(3)}[/tex] = [tex]\sqrt{4}[/tex] × [tex]\sqrt{3}[/tex] = 2[tex]\sqrt{3}[/tex]

Then

[tex]\sqrt{27}[/tex] + [tex]\sqrt{48}[/tex] - [tex]\sqrt{12}[/tex]

= 3[tex]\sqrt{3}[/tex] + 4[tex]\sqrt{3}[/tex] - 2[tex]\sqrt{3}[/tex]

= 5[tex]\sqrt{3}[/tex]

manissaha129

Answer:

[tex] \sqrt{27} + \sqrt{48} - \sqrt{12} \\ \sqrt{3 \times 3 \times 3} + \sqrt{2 \times 2 \times 2 \times 2 \times 3} - \sqrt{2 \times 2 \times 3} \\ 3 \sqrt{3} + 4 \sqrt{3} - 2 \sqrt{3} \\ 7 \sqrt{3} - 2 \sqrt{3} \\ \boxed{5 \sqrt{3}} [/tex]

5√3 is the right answer.

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