1: \(A=\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{x-4}\)
Bài 1:
\(1,A=\dfrac{\sqrt{x}+2+x-9-x+4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\dfrac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ A=\dfrac{\sqrt{x}+1}{x-4}\\ 2,\dfrac{1}{A}\le-\dfrac{5}{2}\Leftrightarrow\dfrac{x-4}{\sqrt{x}+1}+\dfrac{5}{2}\le0\\ \Leftrightarrow\dfrac{x+\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}\le0\\ \Leftrightarrow x+\sqrt{x}-3\le0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow\dfrac{-1-\sqrt{13}}{2}\le\sqrt{x}\le\dfrac{-1+\sqrt{13}}{2}\\ \Leftrightarrow\dfrac{7+\sqrt{13}}{2}\le x\le\dfrac{7-\sqrt{13}}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)
Bài 2:
\(1,P=\dfrac{x\sqrt{x}-x+\sqrt{x}-x\sqrt{x}-x+\sqrt{x}+1-4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{x-\sqrt{x}+1}{\sqrt{x}+11}\\ P=\dfrac{-2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+11\right)}=\dfrac{-2\sqrt{x}}{\sqrt{x}+11}\\ 2,P=\dfrac{-2\left(\sqrt{x}+11\right)+22}{\sqrt{x}+11}=-2+\dfrac{22}{\sqrt{x}+11}\in Z\\ \Leftrightarrow\sqrt{x}+11\inƯ\left(22\right)=\left\{22\right\}\left(\sqrt{x}+11\ge11\right)\\ \Leftrightarrow x=121\)
Bài 3:
\(1,P=\dfrac{\sqrt{a}-1+3\sqrt{a}+5}{\left(a-1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{a-2\sqrt{a}+1}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\\ 2,a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\\ a=\left(8+2\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\\ a=\left(8+2\sqrt{15}\right)\left(8-2\sqrt{15}\right)=64-60=4\\ \Leftrightarrow P=\dfrac{1}{2}\)
