Question

CHO BIỂU THỨC

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

{với x khác 9 ;x lớn hơn hoặc bằng 0}

 

Answer 1

\(ĐKXĐ:\hept{\begin{cases}x\ne9\\x\ge0\end{cases}}\)

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

\(\Leftrightarrow B=\frac{3+\sqrt{x}}{9-x}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{9-x}+\frac{x+9}{9-x}\)

\(\Leftrightarrow B=\frac{3+\sqrt{x}+3\sqrt{x}-x+x+9}{9-x}\)

\(\Leftrightarrow B=\frac{4\sqrt{x}+12}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\)

\(\Leftrightarrow B=\frac{4\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow B=\frac{4}{3-\sqrt{x}}\)