Question

√3(√12+ √27-√48) plz ans ​

Answer 1

$\sqrt{3} \times (2 \sqrt{3 } + 3 \sqrt{3} - 4\sqrt{3} ) \\ \sqrt{3} \times (5 \sqrt{3} - 4 \sqrt{3} ) \\ \sqrt{3} \times \sqrt{3} = 1$

Answer 2

Answer:

$\sqrt{3}\left(\sqrt{12}+\sqrt{27}-\sqrt{48}\right)=3$

Step-by-step explanation:

$\sqrt{3}\left(\sqrt{12}+\sqrt{27}-\sqrt{48}\right)$

$\sqrt{12}$

$\mathrm{Prime\:factorization\:of\:}12:\quad 2^2\cdot \:3$

$=\sqrt{2^2\cdot \:3}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}}$

$=\sqrt{3}\sqrt{2^2}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a}$

$\gray{\sqrt{2^2}=2}$

$=2\sqrt{3}$

$\sqrt{27}$

$\mathrm{Prime\:factorization\:of\:}27:\quad 3^3$

$=\sqrt{3^3}$

$\gray{\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^b\cdot \:a^c}$

$=\sqrt{3^2\cdot \:3}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}}$

$=\sqrt{3}\sqrt{3^2}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a}$

$\gray{\sqrt{3^2}=3}$

$=3\sqrt{3}$

$\sqrt{48}$

$\mathrm{Prime\:factorization\:of\:}48:\quad 2^4\cdot \:3$

$=\sqrt{2^4\cdot \:3}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}}$

$=\sqrt{3}\sqrt{2^4}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\dfrac{m}{n}}}$

$\gray{\sqrt{2^4}=2^{\dfrac{4}{2}}=2^2}$

$=2^2\sqrt{3}$

$\gray{\mathrm{Refine}}$

$=4\sqrt{3}$

$=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-4\sqrt{3}\right)$

$\gray{\mathrm{Add\:similar\:elements:}\:2\sqrt{3}+3\sqrt{3}-4\sqrt{3}=\sqrt{3}}$

$=\sqrt{3}\sqrt{3}$

$\gray{\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{a}=a}$

$\gray{\sqrt{3}\sqrt{3}=3}$

$=3$