Question

if x=5-√3/5+√3 and y=5+√3/5-√3, show that x²-y²=-10√3/11

Answer 1

$x + y = \frac{5 - \sqrt{3} }{5 + \sqrt{3} } + \frac{5 + \sqrt{3} }{5 - \sqrt{3} } \\ = \frac{ {(5 - \sqrt{3} )}^{2} + {(5 + \sqrt{3} )}^{2} }{(5 - \sqrt{3} )(5 + \sqrt{3} )} \\ = \frac{2( {5}^{2} + {( \sqrt{3} )}^{2} )}{ {5}^{2} - {( \sqrt{3}) }^{2} } = \frac{2(25 + 3)}{25 - 3} \\ = \frac{28}{11 } \\ \\ x - y = \frac{ 5- \sqrt{3} }{5 + \sqrt{3} } - \frac{5 + \sqrt{3} }{5 - \sqrt{3} } \\ = \frac{ {(5 - \sqrt{3} )}^{2} - {(5 + \sqrt{3})}^{2} }{ {5}^{2} - {( \sqrt{3} )}^{2} } \\ = \frac{ 2( - 2 \times 5 \times \sqrt{3}) }{25 - 3} = - \frac{10 \sqrt{3} }{11}$
Hope u like the process
=====================
${x}^{2} - {y}^{2} = (x - y)(x + y) \\ = - \frac{10 \sqrt{3} }{11} \times \frac{28}{11} = - \frac{280 \sqrt{3} }{121}$
Hope this is ur required answer
Proud to help you

Answer 2

Answer:

$LHS\neq RHS$

Step-by-step explanation:

Given :  if $x=5-\frac{\sqrt3}{5}+\sqrt3$ and $y=5+\frac{\sqrt3}{5}-\sqrt3$

To Show : $x^2-y^2=-10\frac{\sqrt3}{11}$

Solution :      

We know that, $x^2+y^2=(x+y)(x-y)$

So, First we find the value of x+y by substituting the value of x and y,

$x + y =\frac{5-\sqrt{3}}{5 +\sqrt{3}}+\frac{5 +\sqrt{3} }{5 -\sqrt{3}}\\\\x+y=\frac{{(5-\sqrt{3} )}^{2}+{(5 +\sqrt{3})}^{2}}{(5 -\sqrt{3})(5 +\sqrt{3})}\\\\x+y=\frac{2( {5}^{2}+{( \sqrt{3})}^{2})}{ {5}^{2}-{(\sqrt{3})}^{2}}\\\\x+y=\frac{2(25+3)}{25 - 3}\\\\x+y=\frac{28}{11}$

The, find the value of x-y,

$x-y= \frac{5-\sqrt{3}}{5+ \sqrt{3}}- \frac{5 + \sqrt{3}}{5 -\sqrt{3}}\\\\x-y= \frac{{(5 -\sqrt{3})}^{2}-{(5+ \sqrt{3})}^{2}}{{5}^{2} - {( \sqrt{3} )}^{2}}\\\\x-y = \frac{2( - 2\times 5 \times \sqrt{3})}{25- 3}\\\\x-y=-\frac{10\sqrt{3}}{11}$

Now, Substituting the (x+y) and (x-y) in the formula,

$x^2-y^2=-\frac{10\sqrt{3}}{11}\times\frac{28}{11}\\\\x^2-y^2=-\frac{280\sqrt{3}}{121}$

Since from the result we get $LHS\neq RHS$.