Question

Se considera numerele reale x=$\sqrt{10}$$(\frac{2}{\sqrt{2} }$+$\frac{3}{\sqrt{5} } )$+|3$\sqrt{2}$-6| si y=4+$\frac{10}{\sqrt{5} }$+($\sqrt{3}$)². a)Aratati ca x=2$\sqrt{5}$+6. b)Calculati $(x-y)^{2020}$.

Answer 1

A.

$x = \sqrt{10} ( \frac{2}{ \sqrt{2} } + \frac{3}{ \sqrt{5} } ) + |3 \sqrt{2} - 6 |</p><p>$

$x = 2 \sqrt{5} + 3 \sqrt{2} + |3 \sqrt{2} - 6|$

Pentru a deschide modulul, comparam cele doua numere, punandu-le sub radical

$3 \sqrt{2} \leqslant \geqslant 6$

$\sqrt{18} < \sqrt{36}$

$|3 \sqrt{2} - 6 | = 6 - 3 \sqrt{2}$

$x = 2 \sqrt{5} + 3 \sqrt{2} + 6 - 3 \sqrt{2} = 2 \sqrt{5} + 6$

B.

$y = 4 + \frac{10}{ \sqrt{5} } + ( \sqrt{3} ) {}^{2}$

$y = 4 + \frac{10 \sqrt{5} }{5} + 3$

$y = 7 + 2 \sqrt{5}$

$(x - y) {}^{2020 } = (6 + 2 \sqrt{5} - 7 - 2 \sqrt{5}$

$(x - y) {}^{2020} = ( - 1) {}^{2020}$

Puterea este numar par, asadar rezultatul va trebui sa fie pozitiv

$(x - y) {}^{2020} = 1$