Bài 71 : Cho biểu thức : P=((2sqrt (x))/(sqrt (x)+3)+(sqrt (x))/(sqrt (x)-3)-(3x+3)/(x-9)):((2sqrt (x)-2)/(sqrt (x)-3)-1) a. Rút gọn P. b. Tính P khi x=4-2sqrt (3) c. Tìm x để Plt (1)/(2) d. Tìm GTNN của p

<p><p><span style="text-decoration: underline;"><strong>Đ&aacute;p &aacute;n v&agrave; giải th&iacute;ch c&aacute;c bước giải:</strong></span></p></p><p><p>B&agrave;i $71$:</p></p><p><p>$a,P=(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}):(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1)(x\geq0,x\neq9)$</p></p><p><p>$=(\dfrac{2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3x+3}{x-9}):(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3})$</p></p><p><p>$=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{(\sqrt{x}-3)(\sqrt{x}+3)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}$</p></p><p><p>$=\dfrac{-3\sqrt{x}-3}{(\sqrt{x}-3(\sqrt{x}+3}:\dfrac{\sqrt{x}-1}{\sqrt{x}-3}$</p></p><p><p>$=\dfrac{-3(\sqrt{x}+1)}{(\sqrt{x}-3)(\sqrt{x}+3)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}$</p></p><p><p>$=\dfrac{-3}{\sqrt{x}+3}$</p></p><p><p>____________________________________________________</p></p><p><p>$b,x=4-2\sqrt{3}=3-2\sqrt{3}+1=(\sqrt{3}-1)^{2}$</p></p><p><p>$&rArr;\sqrt{x}=\sqrt{(\sqrt{3}-1)^{2}}=|\sqrt{3}-1|=\sqrt{3}-1$ (v&igrave; $\sqrt{3}&gt;1$)</p></p><p><p>$&rArr;P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{\sqrt{3}+2}$</p></p><p><p>____________________________________________________</p></p><p><p>$c,P&lt;-\dfrac{1}{2}&hArr;\dfrac{-3}{\sqrt{x}+3}&lt;\dfrac{-1}{2}$</p></p><p><p>$&hArr;\dfrac{3}{\sqrt{x}+3}-\dfrac{1}{2}&gt;0$</p></p><p><p>$&hArr;\dfrac{3.2}{2(\sqrt{x}+3)}-\dfrac{\sqrt{x}+3}{2(\sqrt{x}+2}&gt;0$</p></p><p><p>$&hArr;\dfrac{6-\sqrt{x}-3}{2\sqrt{x}+6}&gt;0$</p></p><p><p>$&hArr;\dfrac{3-\sqrt{x}}{2\sqrt{x}+6}&gt;0$</p></p><p><p>\(&hArr;\left[ \begin{array}{l}\left \{ {{3-\sqrt{x}&gt;0} \atop {2\sqrt{x}+6&gt;0}} \right.\\\left \{ {{3-\sqrt{x}&lt;0} \atop {2\sqrt{a}+6&lt;0}} \right.\end{array} \right.\) \(&hArr;\left[ \begin{array}{l}\left \{ {{\sqrt{x}&lt;3} \atop {\sqrt{x}&gt;-3}} \right.\\\left \{ {{\sqrt{x}&gt;3} \atop {\sqrt{x}&lt;-3}} \right.\end{array} \right.\) \(&hArr;\left[ \begin{array}{l}\left \{ {{x&lt;9} \atop {\forall x}} \right.\\\left \{ {{\sqrt{x}&gt;9} \atop {\emptyset}} \right.\end{array} \right.\) $&hArr;x&lt;9$</p></p><p><p>Kết hợp điều kiện x&aacute;c định ta c&oacute;: $0\leq x&lt;9$&nbsp;</p></p><p><p>_______________________________________________________</p></p><p><p>$d,$ Với $x\geq0,x\neq9$ ta c&oacute;: $\sqrt{x}+3\geq3$&nbsp;</p></p><p><p>$&rArr;P=\dfrac{-3}{\sqrt{x}+3}\leq\dfrac{-3}{3}=-1$</p></p><p><p>$&hArr;GTNN:P=-1$</p></p><p><p></p></p><p><p></p></p>