[Toán 9] Căn bặc hai

B

baochauhn1999

$A=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$
$<=>A=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(2+\sqrt{3})^2}}}$
$<=>A=\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}$
$<=>A=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}$
$<=>A=\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}$
$<=>A=\sqrt{5\sqrt{3}+5(5-\sqrt{3})}$
$<=>A=\sqrt{5\sqrt{3}+25-5\sqrt{3}}$
$<=>A=5$
 
B

baochauhn1999

$C=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$
ĐKXĐ: $x$\geq$2$
$=>C\sqrt{2}=\sqrt{2x+4\sqrt{2x-4}}+\sqrt{2x-4\sqrt{2x-4}}$
$<=>C\sqrt{2}=\sqrt{2x-4+4\sqrt{2x-4}+4}+\sqrt{2x-4-4\sqrt{2x-4}+4}$
$<=>C\sqrt{2}=\sqrt{(\sqrt{2x-4}+2)^2}+\sqrt{(\sqrt{2x-4}-2)^2}$
TH1:$\sqrt{2x-4}$\geq$2$
$<=>C\sqrt{2}=\sqrt{2x-4}+2+\sqrt{2x-4}-2=2\sqrt{2x-4}$
$<=>C=2\sqrt{x-2}$
TH2:$\sqrt{2x-4}$\leq$2$
$<=>C\sqrt{2}=\sqrt{2x-4}+2-\sqrt{2x-4}+2=4$
$<=>C=2\sqrt{2}$
 
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