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Áp dụng BĐT Bunhiacopxki, ta có:
$Q=$$\sqrt{x+3}$$+$$\sqrt{10-x}$
$\Leftrightarrow$ $Q^2=($$\sqrt{x+3}$$+$$\sqrt{10-x}$$)^2$ $\le$ $(1^2+1^2)[($
$\sqrt{x+3}$$)^2$$+($$\sqrt{10-x}$$)^2$$]$
$\Leftrightarrow$ $Q^2$ $\le$ $2(x+3+10-x)=2.13=26$
$\Leftrightarrow$$Q$ $\le$ $\sqrt{26}$
$Q_m$$_a$$_x$$=$$\sqrt{26}$ $\Leftrightarrow$ $x+3=10-x$ $\Leftrightarrow$ $x=$$\dfrac{7}{2}$
@Nobitao@
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