Question

Se considera numerele reale a=1 supra √5+2+1 supra 3+√8 si b= 1 supra √5-2+1 supra 3-√8.
Aratati ca a+b=6+2√5.

Answer 1

[tex]a= \frac{1}{ \sqrt{5}+2 }+ \frac{1}{3+ \sqrt{8} } = \frac{1}{ \sqrt{5}+2 }+ \frac{1}{3+ 2\sqrt{2} } \\ b=\frac{1}{ \sqrt{5}-2 }+ \frac{1}{3-\sqrt{8} } =\frac{1}{ \sqrt{5}-2 }+ \frac{1}{3- 2\sqrt{2} } \\ a+b=\frac{1}{ \sqrt{5}+2 }+ \frac{1}{3+ 2\sqrt{2} }+\frac{1}{ \sqrt{5}-2 }+ \frac{1}{3- 2\sqrt{2} }= \\ =(\frac{1}{ \sqrt{5}+2 }+ \frac{1}{ \sqrt{5}-2 })+(\frac{1}{3+ 2\sqrt{2} }+\frac{1}{3- 2\sqrt{2} })=[/tex]
[tex]=(\frac{ \sqrt{5}-2+ \sqrt{5}+2}{(\sqrt{5}+2)(\sqrt{5}-2)})+( \frac{3-2 \sqrt{2}+3+2 \sqrt{2} }{(3+2 \sqrt{2})(3-2 \sqrt{2} ) } )= \frac{2 \sqrt{5} }{( \sqrt{5} )^2-2^2}+ \frac{6}{3^2-(2 \sqrt{2})^2}= \\ = \frac{2 \sqrt{5} }{5-4}+ \frac{6}{9-8}=2 \sqrt{5}+6=6+ 2\sqrt{5} [/tex]
ceea ce trebuia sa demonstram