Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 10.\\ a.\ B=\frac{\sqrt{x} -4}{\sqrt{x}}\\ b.\ AB >2\\ 11.\\ a.\ B=\frac{\sqrt{x} -1}{\sqrt{x} +2}\\ b.AB< 2\\ 12.\\ T< 1\\ 13.\\ M< 1 \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} 10.\\ a.\ B=\frac{\left(\sqrt{x} -1\right)\sqrt{x} -5\sqrt{x} +8}{\sqrt{x}\left(\sqrt{x} -2\right)} =\frac{x-\sqrt{x} -5\sqrt{x} +8}{\sqrt{x}\left(\sqrt{x} -2\right)}\\ =\frac{\left(\sqrt{x} -4\right)\left(\sqrt{x} -2\right)}{\sqrt{x}\left(\sqrt{x} -2\right)} =\frac{\sqrt{x} -4}{\sqrt{x}}\\ b.\ AB-2=\frac{x+\sqrt{x} +1}{\sqrt{x}} -2=\frac{x-\sqrt{x} +1}{\sqrt{x}} >0\ \left( vì\ x-\sqrt{x} +1 >0\right)\\ \Rightarrow AB >2\\ 11.\\ a.\ B=\frac{x+12+\sqrt{x} -2-4\sqrt{x} -2.4}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)} =\frac{x -3\sqrt{x} +2}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)}\\ =\frac{\left(\sqrt{x} -2\right)\left(\sqrt{x} -1\right)}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)} =\frac{\sqrt{x} -1}{\sqrt{x} +2}\\ b.\ AB-2=\frac{\sqrt{x} -1}{\sqrt{x} +2} .\frac{\sqrt{x} +2}{\sqrt{x} +1} -2=\frac{\sqrt{x} -1}{\sqrt{x} +1} -2=\frac{\sqrt{x} -1-2\sqrt{x} -2}{\sqrt{x} +1}\\ =\frac{-\sqrt{x} -3}{\sqrt{x} +1} =\frac{-\left(\sqrt{x} +3\right)}{\sqrt{x} +1} < 0\ \\ \Rightarrow AB< 2\\ 12.\\ T=\left(\frac{a}{4} +\frac{1}{4a} -2.\frac{1}{4}\right) .\frac{\left(\sqrt{a} -1\right)^{2} -\left(\sqrt{a} +1\right)^{2}}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\sqrt{a}\\ =\frac{a^{2} +1-2a}{4a} .\frac{a-2\sqrt{a} +1-a-2\sqrt{a} -1}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\sqrt{a}\\ =\frac{( a-1)^{2}}{4a} .\frac{-4\sqrt{a}}{\left(\sqrt{a} -1\right)\left(\sqrt{a} +1\right)} .\sqrt{a}\\ =1-a< 1\ ( \forall a >0;\ a\neq 1\ )\\ \Rightarrow T< 1\\ 13.\\ M=\frac{\sqrt{a} +1}{\sqrt{a}\left(\sqrt{a} -1\right)} .\frac{\left(\sqrt{a} -1\right)^{2}}{\sqrt{a} +1} =\frac{\sqrt{a} -1}{\sqrt{a}}\\ Ta\ có:\ M-1=\frac{\sqrt{a} -1}{\sqrt{a}} -1=\frac{\sqrt{a} -1-\sqrt{a}}{\sqrt{a}} =\frac{-1}{\sqrt{a}} < 0\\ \Rightarrow M< 1 \end{array}$
