[toán 9] nâng cao

T

thaotran19

Đặt $A=\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}$
=>$A^2=(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2$
$=2\sqrt{5}+2\sqrt{\sqrt{5}+2}.\sqrt{\sqrt{5}-2}$
$=2\sqrt{5}+2.1$
$=2(\sqrt{5}+1)$
=>$A=\sqrt{2(\sqrt{5}+1)}$
=>$\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$
$=\dfrac{\sqrt{2(\sqrt{5}+1)}}{\sqrt{\sqrt{5}+1}}-\sqrt{2-2\sqrt{2}.\sqrt{1}+1}$
$=\sqrt{2}-(\sqrt{2}-1)$

$=1$
 
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