Respuesta :
Answer:
[tex]\frac{-1}{8+5\sqrt{3}}[/tex]The expression we have is:
[tex]\frac{\sqrt{3}-2}{\sqrt{12}+1}[/tex]To rationalize the numerator we must do as follows:
[tex]\frac{\sqrt{3}-2}{\sqrt{12}+1}\cdot\frac{\sqrt{3}+2}{\sqrt{3}+2}[/tex]We multiply the whole expression by the numerator, but we change the sign.
Next, we combine the two fractions:
[tex]\frac{(\sqrt{3}-2)(\sqrt{3}+2)}{(\sqrt{12}+1)(\sqrt{3}+2)}[/tex]Now we use the distributive property to multiply each term (this is to multiply each term in each parenthesis by each term in the other parenthesis besides them).
[tex]\frac{\sqrt{3}\cdot\sqrt{3}+2\sqrt{3}-2\sqrt{3}-2\cdot2}{\sqrt{12}\cdot\sqrt{3}+2\sqrt{12}+1\cdot\sqrt{3}+1\cdot2}[/tex]We simplify the multiplications, and cancel the two middle terms in the numerator:
[tex]\begin{gathered} \frac{\sqrt{3\cdot3}-4}{\sqrt{12\cdot3}+2\sqrt{12}+\sqrt{3}+2} \\ \end{gathered}[/tex]we simplify again:
[tex]\frac{\sqrt{9}-4}{\sqrt{36}+2\sqrt{12}+\sqrt{3}+2}[/tex]We solve the square roots of 9 (which is 3) and 36 (which is 6)
[tex]\frac{3-4}{6+2\sqrt{12}+\sqrt{3}+2}[/tex]Solving the numerator
[tex]\frac{-1}{6+2\sqrt{12}+\sqrt{3}+2}[/tex]And finally what we can do simplify further is to express the square root of 12 as follows:
[tex]\sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3}[/tex]Substituting this into our expression:
[tex]\begin{gathered} \frac{-1}{6+2(2\sqrt{3})+\sqrt{3}+2} \\ \\ \frac{-1}{6+4\sqrt{3}+\sqrt{3}+2} \end{gathered}[/tex]We add 4 and 1 square roots of 3 in the denominar and get 5 square root of 3:
[tex]\frac{-1}{8+5\sqrt{3}}[/tex]also we added 6+2 which is 8.
That is the simplified answer.
