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Đáp án:
a) $B=\dfrac{-3\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}$
b) $A=5$ khi $a=16$
c) $P=-\dfrac{\sqrt{a}+3}{3}$
d) $P\le1$
Giải thích các bước giải:
a)
$B=\dfrac{2\sqrt{a}}{\sqrt{a}+3}-\dfrac{\sqrt{a}}{3-\sqrt{a}}-\dfrac{3a+3}{a-9}\,\,\,(a\ge0;a\ne9)\\=\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3a+3}{(\sqrt{a}-3)(\sqrt{a}+3)}\\=\dfrac{2\sqrt{a}(\sqrt{a}-3)+\sqrt{a}(\sqrt{a}+3)-3a-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\\=\dfrac{2a-6\sqrt{a}+a+3\sqrt{a}-3a-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\\=\dfrac{-3\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}$
b)
$A=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\,\,\,(a\ge0;a\ne9)$
Khi $a=16$
$\to A=\dfrac{\sqrt{16}+1}{\sqrt{16}-3}=\dfrac{4+1}{4-3}=5$
Vậy $A=5$ khi $a=16$
c)
$A=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}; B=\dfrac{-3\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\,\,\,(a\ge0;a\ne9)\\\to P=\dfrac{A}{B}\\=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}:\dfrac{-3\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}\\=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}.\dfrac{(\sqrt{a}-3)(\sqrt{a}+3)}{-3(\sqrt{a}+1)}\\=-\dfrac{\sqrt{a}+3}{3}$
d)
$P=-\dfrac{\sqrt{a}+3}{3}\,\,\,(a\ge0;a\ne9)$
Xét $P-1$
$\to-\dfrac{\sqrt{a}+3}{3}-1\\=\dfrac{-\sqrt{a}-3-3}{3}\\=\dfrac{-(\sqrt{a}+6)}{3}$
Vì $\sqrt{a}+6>0\,\,\,\forall a\ge0\to-(\sqrt{a}+6)\le0$
$\to\dfrac{-(\sqrt{a}+6)}{3}\le0\\\to P-1\le0\\\to P\le1$
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