Question

if x= 4 / 2√3+3√2 then find x+1/x​

Answer 1

Answer:

$\frac{(-10\sqrt{3}+33\sqrt{2})}{12}$

Step-by-step explanation:

$x=\frac{4}{2\sqrt{3}+3\sqrt{2}}$

Rationalise it...

$x =\frac{4}{2\sqrt{3}+3\sqrt{2}  } * \frac{2\sqrt{3} - 3\sqrt{2} }{2\sqrt{3} - 3\sqrt{2}}\\x = \frac{4*(2\sqrt{3} - 3\sqrt{2})}{(2\sqrt{3})^{2} - (3\sqrt{2})^{2}} \\x = \frac{4*(2\sqrt{3} - 3\sqrt{2}}{12 - 18}\\x = \frac{4*(2\sqrt{3} - 3\sqrt{2}}{-6}\\x = \frac{2*(2\sqrt{3} - 3\sqrt{2}}{-3}\\x = \frac{4\sqrt{3}-6\sqrt{2}}{-3}$

Now find the value of 1/x

$\frac{1}{x} =\frac{1}{\frac{4\sqrt{3}-6\sqrt{2}}{-3}} \\\frac{1}{x} =\frac{-3}{4\sqrt{3}-6\sqrt{2}}\\\frac{1}{x} =\frac{-3}{4\sqrt{3}-6\sqrt{2}}*\frac{4\sqrt{3}+6\sqrt{2}}{4\sqrt{3}+6\sqrt{2}}\\\frac{1}{x} =\frac{-3*(4\sqrt{3}+6\sqrt{2}}{-24}\\\frac{1}{x} =\frac{(4\sqrt{3}+6\sqrt{2}}{8}$

now add : x + 1/x

$\frac{4\sqrt{3}-6\sqrt{2}}{-3}+\frac{4\sqrt{3}+6\sqrt{2}}{8}\\LCM = 24\\\frac{8*(-4\sqrt{3}+6\sqrt{2})}{3*8}+\frac{3*(4\sqrt{3}+6\sqrt{2})}{8*3}\\\frac{-32\sqrt{3}+48\sqrt{2})}{24}+\frac{12\sqrt{3}+18\sqrt{2})}{24}\\\frac{-20\sqrt{3}+66\sqrt{2}}{24} \\\frac{2*(-10\sqrt{3}+33\sqrt{2})}{24}\\\frac{(-10\sqrt{3}+33\sqrt{2})}{12} ANSWER$

MARK BRAINLIEST

THANKS