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\(a,A=\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{3}{\sqrt{x}+2}-\frac{9\sqrt{x}-10}{x-4}\left(x\ge0;x\ne16\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{3\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{9\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}+3\sqrt{x}-6-9\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}\right)^2-2.\sqrt{x}.2+2^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
Vây...
\(b,\)Ta có:\(x=4-2\sqrt{3}=\left(1-\sqrt{3}\right)^2\)
Thay \(x=\left(1-\sqrt{3}\right)^2\)vào A ta được:
\(A=\frac{\sqrt{\left(1-\sqrt{3}\right)^2}-2}{\sqrt{\left(1-\sqrt{3}\right)^2}+2}=\frac{\sqrt{3}-1-2}{\sqrt{3}-1+2}=\frac{\sqrt{3}-3}{\sqrt{3}-1}=\frac{-\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}=-\sqrt{3}\)