Math, asked by ab12, 1 year ago

If x =2√15/√5+√3, find x+√3/x-√3 + x+√5/x-√5


AvmnuSng: write question clearly ....

Answers

Answered by chandratej007
16
x+√3/x-√3    +    x+√5/x-√5
 substitute the value of "x"
=> 2√15/√5+√3 +√3/2√15/√5+√3-√3          +        2√15/√5+√3+√5/2√15/√5+√3-√5

2√15+√15+3/√5+√3/2√15-√15-3/√5+√3 +  2√15+5+√15/√5+√3/2√15-5-            √15/√5+√3  
 
3√15+3/√15-3     +    3√15+5/√15-5

(√15-5)(3√15+3) + (√15-3)(3√15+5)/(√15-3)(√15-5)

45-15+3√15-15√15+45-15+5√15-9√15/30-5√15-3√15

60-16√15/30-8√15

2(30-8√15)/30-8√15

then the answer we get is        2       
     





Answered by kvnmurty
55
x = \frac{2\sqrt{15}}{\sqrt{5}+\sqrt{3}} = \frac{2\sqrt{15}(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} = \frac{2\sqrt{75} - 2\sqrt{45}}{5-3} = 5\sqrt{3}-3\sqrt5 \\ \\ \\ \frac{x + \sqrt{3}}{x - \sqrt{3}} + \frac{x+\sqrt5}{x-\sqrt5} \\ \\ , \ \ \ \ Rationalize\ each\ fraction. \\ \\

[tex]\frac{(6\sqrt3-3\sqrt5)(4\sqrt3+3\sqrt5)}{(4\sqrt3-3\sqrt5)(4\sqrt3+3\sqrt5)} + \frac{(5\sqrt3-2\sqrt5)(5\sqrt3 + 4\sqrt5)}{(5\sqrt3 - 4\sqrt5)(5\sqrt3 + 4\sqrt5)} \\ \\ \frac{24*3+18\sqrt{15}-12\sqrt{15}-3^2*5}{4^2*3 - 3^2*5} + \frac{5^2*3+20\sqrt{15}-10\sqrt{15}-8*5}{25*3 - 16*5} \\ \\ \frac{27+6\sqrt{15}}{3} + \frac{35+10\sqrt{15}}{-5} \\ \\ 9+2\sqrt{15}-7-2\sqrt{15} \\ \\ 2 \\ [/tex]

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