Question

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

Tìm đkxđ của biểu thức A

Rút gọn A

Answer 1

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

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

\(Đkxđ:\)

\(\sqrt{x}\ge0\Rightarrow x\ge0\)

\(\sqrt{x}-1\ne0\Rightarrow\sqrt{x}\ne1\Rightarrow x\ne1\)

\(\sqrt{x}\ne0\Rightarrow x\ne0\)

\(\RightarrowĐkxđ:x>0;x\ne1\)

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

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

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

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

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

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

Answer 2

\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}}\)

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

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

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

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

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