Question

\(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3}{3-\sqrt{x}}-\frac{6\sqrt{x}}{x-9}\)

Answer 1

\(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3}{3-\sqrt{x}}+\frac{6\sqrt{x}}{9-x}\)

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

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

\(=\frac{-x+6\sqrt{x}-9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\)

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

Answer 2

nâu thanks ak

Answer 3

\(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3}{3-\sqrt{x}}-\frac{6\sqrt{x}}{x-9}\)

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

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

=\(\frac{x-3\sqrt{x}+3\sqrt{x}+9-6\sqrt{x}}{\left(\sqrt{x+3}\right)\left(\sqrt{x}-3\right)}\)

=\(\frac{x+9-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

=\(\frac{x-6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

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