Question

maths
Simplify:6÷√12-√3​

Answer 1

Given :-

$\sf \bullet \;\; \dfrac{6}{\sqrt{12} -\sqrt{3}}$

To Do :-

• Simplify

Solution :-

$\sf : \; \implies \dfrac{6}{\sqrt{3 \times 2 \times 2} - \sqrt{3} }$

$\sf : \; \implies \dfrac{6}{ 2\sqrt{3} - \sqrt{3} }$

$\sf : \; \implies \dfrac{6}{ \sqrt{3} }$

$\sf : \; \implies \dfrac{6}{ \sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} }$

$\sf : \; \implies \dfrac{6\sqrt{3} }{ (\sqrt{3})^{2} }$

$\sf : \; \implies \dfrac{6\sqrt{3} }{ 3}$

$\boxed{\bf{ \star \;\; 2 \sqrt{3} \;\; }}$

Answer 2

Given :-

6 ÷ √12-√3​

To Find :-

Simplified Value

Solution :-

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀

• Factor 12 = 2²×3. Rewrite the square root of the product √2² × 3    as the product of square roots √2² × √3. Take the square root of 2².

$\sf \dashrightarrow \; \dfrac{6}{2\sqrt{3} - \sqrt{3}}$

• Combine 2√3 and -√3 to get √3

$\sf \dashrightarrow \; \dfrac{6}{ \sqrt{3}}$

• Rationalize the denominator of √3 by multiplying numerator and denominator by √3.

$\sf \dashrightarrow \; \bigg( \dfrac{6 \sqrt{3} }{ ( \sqrt{3} )^{2} }\bigg)$

$\sf \dashrightarrow \; \dfrac{6 \sqrt{3} }{ 3 }$

$\sf \dashrightarrow 2\sqrt{3}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀

• Henceforth, the simplified value is 2√3.