Question

find the value ....​

Answer 1

Answer:

$\sqrt{75}+\dfrac{1}{2}\sqrt{48}-\sqrt{192}=-\sqrt{3}\quad\mathrm{or}\quad-1.732$

Step-by-step explanation:

$\sqrt{75}+\dfrac{1}{2}\sqrt{48}-\sqrt{192}$

$\black{\sqrt{75}}$

$\gray{\mathrm{Prime\:factorization\:of\:}75:\quad 3\cdot \:5^2}$

$=\sqrt{5^2\cdot \:3}$

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

$=\sqrt{3}\sqrt{5^2}$

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

$\gray{\sqrt{5^2}=5}$

$=5\sqrt{3}$

$\black{\dfrac{1}{2}\sqrt{48}}$

$\gray{\sqrt{48}=4\sqrt{3}}$

$=4\sqrt{3}\dfrac{1}{2}$

$\gray{\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}}$

$=\dfrac{1\cdot \:4\sqrt{3}}{2}$

$\gray{\mathrm{Multiply\:the\:numbers:}\:1\cdot \:4=4}$

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

$\gray{\mathrm{Divide\:the\:numbers:}\:\dfrac{4}{2}=2}$

$=2\sqrt{3}$

$\black{\sqrt{192}}$

$\gray{\mathrm{Prime\:factorization\:of\:}192:\quad 2^6\cdot \:3}$

$=\sqrt{2^6\cdot \:3}$

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

$=\sqrt{3}\sqrt{2^6}$

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

$\gray{\sqrt{2^6}=2^{\dfrac{6}{2}}=2^3}$

$=2^3\sqrt{3}$

$\gray{\mathrm{Refine}}$

$=8\sqrt{3}$

$=5\sqrt{3}+2\sqrt{3}-8\sqrt{3}$

$\gray{\mathrm{Add\:similar\:elements:}\:5\sqrt{3}+2\sqrt{3}-8\sqrt{3}=-\sqrt{3}}$

$=-\sqrt{3}$

$= - (\sqrt{3})$

$=-(1.732)$

$=-1.732$