ĐKXĐ: x\(\ge\)3 x\(\ne4\)
B=\(\dfrac{\sqrt{x-2-2\sqrt{x-3}}}{\sqrt{x-3}-1}=\dfrac{\sqrt{x-3-2\sqrt{x-3}+1}}{\sqrt{x-3}-1}=\dfrac{\sqrt{\left(\sqrt{x-3}-1\right)^2}}{\sqrt{x-3}-1}=\dfrac{\left|\sqrt{x-3}-1\right|}{\sqrt{x-3}-1}\)
Nếu x>4 =>\(\sqrt{x-3}-1>0\Rightarrow B=1\)
Nếu \(3\le x< 4\)=>\(\sqrt{x-3}-1< 0\Rightarrow B=-1\)
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