Đáp án:
$a)\text{ĐKXĐ: } \left\{\begin{array}{l} x \ge 0 \\x \ne 1\end{array} \right.\\ A=\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ b)A(25)=\dfrac{5}{6}\\ c)x=4\\ d)x=0$
Giải thích các bước giải:
$A=\dfrac{x-\sqrt{x}-2}{x-1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\\ a)\text{ĐKXĐ: } \left\{\begin{array}{l} x \ge 0 \\x-1 \ne 0 \\ \sqrt{x}-1 \ne 0 \\ \sqrt{x}+1 \ne 0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\x \ne 1 \\ \sqrt{x}\ne 1 \\ \sqrt{x} \ne -1\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\x \ne 1\end{array} \right.\\ A=\dfrac{x-\sqrt{x}-2}{x-1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\\ =\dfrac{x-\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\\ =\dfrac{x-\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}+\dfrac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{x-\sqrt{x}-2+\sqrt{x}+1-\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{x-\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ b)A(25)=\dfrac{\sqrt{25}}{\sqrt{25}+1}=\dfrac{5}{5+1}=\dfrac{5}{6}\\ c)A=\dfrac{2}{3}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{2}{3}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{3}=0\\ \Leftrightarrow \dfrac{3\sqrt{x}-2(\sqrt{x}+1)}{3(\sqrt{x}+1)}=0\\ \Leftrightarrow \dfrac{\sqrt{x}-2}{3(\sqrt{x}+1)}=0\\ \Leftrightarrow \sqrt{x}-2=0\\ \Leftrightarrow \sqrt{x}=2\\ \Leftrightarrow x=4\\ d)A=\dfrac{\sqrt{x}}{2}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{2}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{2}=0\\ \Leftrightarrow \dfrac{2\sqrt{x}-\sqrt{x}(\sqrt{x}+1)}{2(\sqrt{x}+1)}=0\\ \Leftrightarrow \dfrac{\sqrt{x}-x}{2(\sqrt{x}+1)}=0\\ \Leftrightarrow \sqrt{x}-x=0\\ \Leftrightarrow \sqrt{x}(\sqrt{x}-1)=0\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=0\\ \sqrt{x}=1\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=0\\ x=1(L)\end{array} \right.\\ e)C=A.\dfrac{x+3}{\sqrt{x}}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{x+3}{\sqrt{x}}\\ =\dfrac{x+3}{\sqrt{x}+1}\\ =\dfrac{x+2\sqrt{x}+1-2\sqrt{x}-2+4}{\sqrt{x}+1}\\ =\dfrac{(\sqrt{x}+1)^2-2(\sqrt{x}+1)+4}{\sqrt{x}+1}\\ =\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\\ \ge 2\sqrt{(\sqrt{x}+1)\dfrac{4}{\sqrt{x}+1}}-2=2$
Dấu "=" xảy ra
$\Leftrightarrow \sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}\Leftrightarrow (\sqrt{x}+1)^2=4\Leftrightarrow \left[\begin{array}{l} \sqrt{x}+1=2 \\ \sqrt{x}+1=-2\end{array} \right.\Leftrightarrow x=1(L)$
Vậy C không tồn tại GTNN.
