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\(\sqrt{\left(6-2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}=0\)
\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{2}.\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{2}.\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\sqrt{\left(\sqrt{5}+1\right)^2}+\left(3+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-1^2\right)}\)
\(\Leftrightarrow\sqrt{2B}=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left|\sqrt{5}-1\right|\)
\(=3\sqrt{5}+3-5-\sqrt{5}+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(=6\sqrt{5}-2\sqrt{5}=4\sqrt{5}\)
\(\Rightarrow B=\frac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
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)
\(B=b^2\cdot a+a^2\cdot b=ab\left(a+b\right)=2\sqrt{10}\)
bài này mình cũng dò lại đề rồi mình chép đúng đấy mà không làm được nên mới nhờ giải
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
\(\sqrt{4+2\sqrt{3}}=2.73\)
\(\left(\sqrt{7}+\sqrt{3}\right)^2=10+2\sqrt{21}\)
\(\left(\sqrt{5}-\sqrt{3}+1\right)\left(\sqrt{5}-1\right)=1.86\)
\(\left(5\sqrt{3}-2\sqrt{7}\right)\left(5\sqrt{3}+2\sqrt{7}\right)=47\)