Question

Calculati:
($\frac{5}{2\sqrt{2}+\sqrt{3} } -\frac{2}{\sqrt{5}+\sqrt{3} } +\frac{18}{2\sqrt{5}-\sqrt{2} } -\frac{3}{\sqrt{5}-2\sqrt{2} }$)*$\frac{\sqrt{10} }{\sqrt{5}+\sqrt{2} }$



Ofer coroana +98 puncte

va rog

Answer 1

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Answer 2

[tex]\it\left(\dfrac{5}{2\sqrt2+\sqrt3}- \dfrac{2}{\ \ \sqrt5+\sqrt3}+\dfrac{18}{2\sqrt5+\sqrt2}- \dfrac{3}{\sqrt5-2\sqrt2}\right)\cdot \dfrac{\sqrt{10}}{\sqrt5+\sqrt2}= \\ \\ \\ = \left(\dfrac{^{2\sqrt2-\sqrt3)}5}{\ \ \ 2\sqrt2+\sqrt3}- \dfrac{^{\sqrt5-\sqrt3)}2}{\sqrt5+\sqrt3}+\dfrac{^{2\sqrt5-\sqrt2)}18}{2\sqrt5+\sqrt2}+ \dfrac{^{2\sqrt2+\sqrt5)}3}{\ \ \ 2\sqrt2-\sqrt5}\right)\cdot \\ \\ \\ \cdot \dfrac{\sqrt{10}}{\sqrt5+\sqrt2} =[/tex]

[tex]\it = \left(\dfrac{5(2\sqrt2-\sqrt3)}{5} -\dfrac{2(\sqrt5-\sqrt3)}{2} + \dfrac{18(2\sqrt5+\sqrt2)}{18}+\dfrac{3(2\sqrt2-\sqrt5)}{3}\right)\cdot \\ \\ \\ \cdot \dfrac{\sqrt{10}}{\sqrt5+\sqrt2} = (2\sqrt2-\sqrt3-\sqrt5+\sqrt3+2\sqrt5+\sqrt2+2\sqrt2- \sqrt5)\cdot \dfrac{\sqrt{10}}{\sqrt5+\sqrt2} = \\ \\ \\ = (5\sqrt2+2\sqrt5)\cdot \dfrac{\sqrt{10}}{\sqrt5+\sqrt2} =[/tex]

$\it = \sqrt2\cdot\sqrt5(\sqrt5+\sqrt2)\cdot\dfrac{\sqrt{10}}{\sqrt5+\sqrt2} =\sqrt{10}\cdot\sqrt{10} =10$