Question

Answer 1

\(a,ĐK:x\ge0;x\ne1\\ b,Q=\dfrac{x+\sqrt{x}+\sqrt{x}-x-3+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\\ Q=\dfrac{3\sqrt{x}-3}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\dfrac{-3\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\dfrac{-3}{\sqrt{x}+1}\\ c,x=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\\ \Leftrightarrow Q=\dfrac{-3}{\sqrt{3}-1+1}=\dfrac{-3}{\sqrt{3}}=-\sqrt{3}\\ d,Q\in Z\\ \Leftrightarrow\sqrt{x}+1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;2\right\}\\ \Leftrightarrow x\in\left\{1;4\right\}\)