4)
a) \(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=\left(\sqrt{5}+2\right)-\left(\sqrt{5}-2\right)=4\)
b) \(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)=-2\)
3. Chứng minh \(6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(6-2\sqrt{5}=\left(\sqrt{5}\right)^2-2.\sqrt{5}.1+1^2=\left(\sqrt{5}-1\right)^2\)
Vậy \(6-2\sqrt[]{5}=\left(\sqrt{5}-1\right)^2\)
\(M=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1-\left(1-\sqrt{5}\right)=\sqrt{5}+1-1+\sqrt{5}=\sqrt{5}\)
\(\left(Đk:\sqrt{5}+1>0,\sqrt{5}-1< 0\right)\)
4.
a) \(M=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(M=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(M=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(M=2+\sqrt{5}-\left(\sqrt{5}-2\right)\) Đk :\(2+\sqrt{5}>0\), \(2-\sqrt{5}< 0\)
\(M=4\)
b) \(N=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(N=\sqrt{1^2-2.1.\sqrt{7}+\left(\sqrt{7}\right)^2}-\sqrt{1^2+2.1.\sqrt{7}+\left(\sqrt{7}\right)^2}\)
\(N=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+\sqrt{7}\right)^2}\)
\(N=\sqrt{7}-1-\left(1+\sqrt{7}\right)\) Đk :\(1-\sqrt{7}< 0\), \(1+\sqrt{7}>0\)
\(N=\sqrt{7}-1-1-\sqrt{7}=-2\)
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Câu 3
\(M=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1-\sqrt{5}+1=2\)
