Question

[Le DuyDucc_]

Giúp mình với

Answer 1

[Nguyễn Huy Tú]

a, Với \(x\ge0;x\ne2\)

\(P=\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}+2}{1-\sqrt{x}}\)

\(=\frac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{3x+3\sqrt{x}-3-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-3}{\sqrt{x}-1}\)

b, \(\frac{\sqrt{x}-3}{\sqrt{x}-1}=\frac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\frac{2}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\sqrt{x}-1\)	1	-1	2	-2
x	4	0	9	loại