giúp em vs gấp quá
a: \(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\frac{-\left(\sqrt{x}+2\right)}{\sqrt{x}+3}\)
\(=\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\frac{\sqrt{x}+3}{-\left(\sqrt{x}+2\right)}=\frac{-3}{-\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)
b: \(A=\frac12\)
=>\(\frac{3}{\sqrt{x}+2}=\frac12\)
=>\(\sqrt{x}+2=6\)
=>\(\sqrt{x}=4\)
=>x=16(nhận)
c: Thay \(x=19-8\sqrt3=\left(4-\sqrt3\right)^2\) vào A, ta được:
\(A=\frac{3}{\sqrt{\left(4-\sqrt3\right)^2}+2}=\frac{3}{4-\sqrt3+2}=\frac{3}{6-\sqrt3}\)
\(=\frac{3\left(6+\sqrt3\right)}{\left(6-\sqrt3\right)\left(6+\sqrt3\right)}=\frac{3\left(6+\sqrt3\right)}{36-3}=\frac{6+\sqrt3}{11}\)
