ĐKXĐ: x>=4
Ta có: \(B=\sqrt{x-3-2\sqrt{x-4}}+\sqrt{x-3+2\sqrt{x-4}}\)
\(=\sqrt{x-4-2\cdot\sqrt{x-4}\cdot1+1}+\sqrt{x-4+2\cdot\sqrt{x-4}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{x-4}-1\right)^2}+\sqrt{\left(\sqrt{x-4}+1\right)^2}=\left|\sqrt{x-4}-1\right|+\sqrt{x-4}+1\)
TH1: \(\sqrt{x-4}-1\ge0\)
=>\(\sqrt{x-4}\ge1\)
=>x-4>=1
=>x>=5
\(B=\left|\sqrt{x-4}-1\right|+\sqrt{x-4}+1\)
\(=\sqrt{x-4}-1+\sqrt{x-4}+1\)
\(=2\sqrt{x-4}\)
TH2: \(\sqrt{x-4}-1<0\)
=>\(\sqrt{x-4}<1\)
=>0<=x-4<1
=>4<=x<5
\(B=\left|\sqrt{x-4}-1\right|+\sqrt{x-4}+1\)
\(=1-\sqrt{x-4}+\sqrt{x-4}+1\)
=2
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