\(\dfrac{x^2-3}{x+\sqrt{3}}=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
Bài 1:
\(a,=\dfrac{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\\ b,=\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}=a+\sqrt{a}+1\)
Bài 2:
\(a,A=\dfrac{a+\sqrt{a}-2-a+\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\left(\sqrt{a}+1\right)\\ A=\dfrac{2\sqrt{a}}{\sqrt{a}-1}\\ b,a=\dfrac{2}{2-\sqrt{3}}=\dfrac{2\left(2+\sqrt{3}\right)}{4-3}=4+2\sqrt{3}\Leftrightarrow\sqrt{a}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{2\left(\sqrt{3}+1\right)}{\sqrt{3}+1-1}=\dfrac{2\sqrt{3}+2}{\sqrt{3}}=\dfrac{6+2\sqrt{3}}{3}\\ c,A=\dfrac{2\left(\sqrt{a}-1\right)+2}{\sqrt{a}-1}=2+\dfrac{2}{\sqrt{a}-1}\in Z\\ \Leftrightarrow\sqrt{a}-1\inƯ\left(2\right)=\left\{-1;1;2\right\}\left(\sqrt{a}-1\ge-1\right)\\ \Leftrightarrow\sqrt{a}\in\left\{0;2;3\right\}\\ \Leftrightarrow a\in\left\{0;4;9\right\}\left(tm\right)\)
