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Đáp án:

$4)\\ a)D=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\\ 5)\\ a)-\sqrt{x}(\sqrt{x}-1)\\ b)max_P=\dfrac{1}{4} \Leftrightarrow x=\dfrac{1}{4}$

Giải thích các bước giải:

$4)\\ a)D=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left[\dfrac{6\sqrt{x}+1}{(2\sqrt{x}-3)(\sqrt{x}+1)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right]\\ =\dfrac{2(2\sqrt{x}-3)-(\sqrt{x}-1)}{2\sqrt{x}-3}:\left[\dfrac{6\sqrt{x}+1}{(2\sqrt{x}-3)(\sqrt{x}+1)}+\dfrac{\sqrt{x}(2\sqrt{x}-3)}{(\sqrt{x}+1)(2\sqrt{x}-3)}\right]\\ =\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}:\dfrac{2x+3\sqrt{x}+1}{(\sqrt{x}+1)(2\sqrt{x}-3)}\\ =\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}.\dfrac{(\sqrt{x}+1)(2\sqrt{x}-3)}{2x+2\sqrt{x}+\sqrt{x}+1}\\ =\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}.\dfrac{(\sqrt{x}+1)(2\sqrt{x}-3)}{(2\sqrt{x}+1)(\sqrt{x}+1)}\\ =\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\\ b)D-\dfrac{3}{4}\\ =\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}-\dfrac{3}{4}\\ =\dfrac{4(3\sqrt{x}-5)-3(2\sqrt{x}+1)}{2\sqrt{x}+1}\\ =\dfrac{6\sqrt{x}-23}{2\sqrt{x}+1} \text{ không nhỏ hơn } 0  \ \forall x \ge 0, x \ne \dfrac{9}{4}\\ \Rightarrow \text{Đề sai}\\ 5)\\ a)P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{(1-x)^2}{2}\\ =\left(\dfrac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2}\right).\dfrac{(1-x)^2}{2}\\ =\left(\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}-\dfrac{(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}\right).\dfrac{(1-x)^2}{2}\\ =\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)-(\sqrt{x}+2)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\dfrac{(1-x)^2}{2}\\ =\dfrac{-2\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)^2}.\dfrac{(\sqrt{x}-1)^2(\sqrt{x}+1)^2}{2}\\ =-\sqrt{x}(\sqrt{x}-1)\\ b)P=-\sqrt{x}(\sqrt{x}-1)\\ =-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}\\ =-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4} \le \dfrac{1}{4} \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}-\dfrac{1}{2}=0 \Leftrightarrow x=\dfrac{1}{4}$