Question

simplify 4 root 3/2-root 2-30/4root3-3root2-3 root 2/3+2 root 3

Answer 1

$\textbf{To simplify:}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}}$

$\mathsf{=\dfrac{4\sqrt{3}}{2-\sqrt{2}}{\times}\dfrac{2+\sqrt{2}}{2+\sqrt{2}}}$

$\mathsf{=\dfrac{4\sqrt{3}(2+\sqrt{2})}{4-2}}$

$\mathsf{=\dfrac{4\sqrt{3}(2+\sqrt{2})}{2}}$

$\mathsf{=2\sqrt{3}(2+\sqrt{2})}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}}$

$\mathsf{\dfrac{30}{4\sqrt{3}-3\sqrt{2}}}$

$\mathsf{=\dfrac{30}{4\sqrt{3}-3\sqrt{2}}{\times}\dfrac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}$

$\mathsf{=\dfrac{30(4\sqrt{3}+3\sqrt{2})}{48-18}}$

$\mathsf{=\dfrac{30(4\sqrt{3}+3\sqrt{2})}{30}}$

$\mathsf{=4\sqrt{3}+3\sqrt{2}}$

$\mathsf{\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\mathsf{=\dfrac{3\sqrt{2}}{3+2\sqrt{3}}{\times}\dfrac{3-2\sqrt{3}}{3-2\sqrt{3}}}$

$\mathsf{=\dfrac{3\sqrt{2}(3-2\sqrt{3})}{9-12}}$

$\mathsf{=\dfrac{3\sqrt{2}(3-2\sqrt{3})}{-3}}$

$\mathsf{=-\sqrt{2}(3-2\sqrt{3})}$

$\mathsf{=-3\sqrt{2}+2\sqrt{6}}$

$\mathsf{Now,}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}-(4\sqrt{3}+3\sqrt{2})-(-3\sqrt{2}+2\sqrt{6})}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{6})}$

$\mathsf{=0}$

$\implies\boxed{\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}=0}}$

$\textbf{Find more:}$

√7 - 2/√7 + 2 = a√7 + b

https://brainly.in/question/1115376

If x = √7+ √3 and xy = 4, then x4 + y4 =​

https://brainly.in/question/16537214

If p and q are rational numbers and

p√15q ? 2√3 -√5/4√3-3√5 find the value of p and q

https://brainly.in/question/16694887

Answer 2

$\textbf{To simplify:}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}}$

$\mathsf{=\dfrac{4\sqrt{3}}{2-\sqrt{2}}{\times}\dfrac{2+\sqrt{2}}{2+\sqrt{2}}}$

$\mathsf{=\dfrac{4\sqrt{3}(2+\sqrt{2})}{4-2}}$

$\mathsf{=\dfrac{4\sqrt{3}(2+\sqrt{2})}{2}}$

$\mathsf{=2\sqrt{3}(2+\sqrt{2})}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}}$

$\mathsf{\dfrac{30}{4\sqrt{3}-3\sqrt{2}}}$

$\mathsf{=\dfrac{30}{4\sqrt{3}-3\sqrt{2}}{\times}\dfrac{4\sqrt{3}+3\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}}$

$\mathsf{=\dfrac{30(4\sqrt{3}+3\sqrt{2})}{48-18}}$

$\mathsf{=\dfrac{30(4\sqrt{3}+3\sqrt{2})}{30}}$

$\mathsf{=4\sqrt{3}+3\sqrt{2}}$

$\mathsf{\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\mathsf{=\dfrac{3\sqrt{2}}{3+2\sqrt{3}}{\times}\dfrac{3-2\sqrt{3}}{3-2\sqrt{3}}}$

$\mathsf{=\dfrac{3\sqrt{2}(3-2\sqrt{3})}{9-12}}$

$\mathsf{=\dfrac{3\sqrt{2}(3-2\sqrt{3})}{-3}}$

$\mathsf{=-\sqrt{2}(3-2\sqrt{3})}$

$\mathsf{=-3\sqrt{2}+2\sqrt{6}}$

$\mathsf{Now,}$

$\mathsf{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}-(4\sqrt{3}+3\sqrt{2})-(-3\sqrt{2}+2\sqrt{6})}$

$\mathsf{=4\sqrt{3}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}-2\sqrt{6})}$

$\mathsf{=0}$

$\implies\boxed{\mathsf\red{\dfrac{4\sqrt{3}}{2-\sqrt{2}}-\dfrac{30}{4\sqrt{3}-3\sqrt{2}}-\dfrac{3\sqrt{2}}{3+2\sqrt{3}}=0}}$