Question

Rút gọn biểu thức:

\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

Answer 1

\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)

\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)

\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1^2\right)}\)

\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)

\(=3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)

\(=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)

\(=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\)

\(\Rightarrow B=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

Answer 2

Đặt \(\sqrt{3+\sqrt{5}}=a>0;\sqrt{3-\sqrt{5}}=b>0\Rightarrow ab=\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\sqrt{3^2-5}=2\)

Và \(a^2+b^2=6 \Rightarrow\left(a+b\right)^2=a^2+b^2+2ab=6+4=10\Rightarrow a+b=\sqrt{10}\) (vì a + b > 0 do a > 0,b>0)

\(B=b^2\cdot a+a^2\cdot b=ab\left(a+b\right)=2\sqrt{10}\)