Respuesta :
Answer:
[tex]6\sqrt{2}+2{\sqrt3}[/tex]
Step-by-step explanation:
Given expression:
[tex]\sqrt{72}-3\sqrt{12}+\sqrt{192}[/tex]
Rewrite 72 as (36 · 2), 12 as (4 · 3), and 192 as (64 · 3):
[tex]\implies \sqrt{36 \cdot 2}-3\sqrt{4 \cdot 3}+\sqrt{64 \cdot 3}[/tex]
Apply the radical rule [tex]\sqrt{a \cdot b}=\sqrt{a}\sqrt{b}[/tex] :
[tex]\implies \sqrt{36}\sqrt{2}-3\sqrt{4}\sqrt{3}+\sqrt{64}{\sqrt3}[/tex]
Rewrite 36 as 6², 4 as 2², and 64 as 8²:
[tex]\implies \sqrt{6^2}\sqrt{2}-3\sqrt{2^2}\sqrt{3}+\sqrt{8^2}{\sqrt3}[/tex]
Apply the radical rule [tex]\sqrt{a^2}=a[/tex] :
[tex]\implies 6\sqrt{2}-3\cdot 2\sqrt{3}+8{\sqrt3}[/tex]
Simplify:
[tex]\implies 6\sqrt{2}-6\sqrt{3}+8{\sqrt3}[/tex]
[tex]\implies 6\sqrt{2}+2{\sqrt3}[/tex]
Answer:
6√2+2√3
Step-by-step explanation:
We want to simplify the following radical expression
[tex] \displaystyle \sqrt{72} - 3 \sqrt{12} + \sqrt{192} [/tex]
Recall that
[tex] \sqrt{ab} = \sqrt{a} \sqrt{b} , \forall \text{a and b such that a$\geq$0,b$\geq$0} [/tex]
Utilizing the formula yields,
[tex] \sqrt{72} \implies \sqrt{36 \cdot 2} \implies 6 \sqrt{2} [/tex]
[tex] \sqrt{12} \implies \sqrt{4\cdot 3} \implies 2 \sqrt{3} [/tex]
[tex] \sqrt{192} \implies \sqrt{64\cdot 3} \implies 8 \sqrt{3} [/tex]
So,
[tex]6 \sqrt{2}-3\cdot2 \sqrt{3}+8 \sqrt{3}[/tex]
Carry out multiplication:
[tex]\implies 6 \sqrt{2}-6 \sqrt{3}+8 \sqrt{3}[/tex]
Add the like terms:
[tex]\boxed{6 \sqrt{2}+2 \sqrt{3}}[/tex]
and we're done!