Question

cho B= \(\frac{x}{x-4}\) - \(\frac{1}{2-\sqrt{x}}\) + \(\frac{1}{2+\sqrt{x}}\)
a) tìm ĐKXĐ
b) rút gọn biểu thức B

Answer 1

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

a)\(ĐKXĐ:\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)

b) Ta có: \(B=\frac{x+\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}\)

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

Vậy \(B=\frac{\sqrt{x}}{\sqrt{x}-2}\)

Answer 2

Lời giải:
a) ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ x-4\neq 0\\ 2-\sqrt{x}\neq 0\\ 2+\sqrt{x}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 4\end{matrix}\right.\)

b)

\(B=\frac{x}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}=\frac{x}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

\(=\frac{x+2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{\sqrt{x}-2}\)