Rút gọn các biểu thức sau: ko tính
\(13-2\sqrt{42}\)
\(46+6\sqrt{5}\)
\(\sqrt{3-\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\sqrt[]{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}\)
\(13-2\sqrt{42}=7-2\sqrt{42}+6\\ =\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{7}-\sqrt{6}\right)^2\)
\(46+6\sqrt{5}=\left(5+2\cdot\sqrt{5}\cdot3+9\right)+32=\left(\sqrt{5}+3\right)^2+32\)(ko rút đc)
\(\sqrt{3-\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\\ =\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{6-2\sqrt{5}}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{5-2\sqrt{5}+1}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\\ =\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\\ =4\left(3+\sqrt{5}\right)\)
\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Dễ dàng nhận ra
\(\sqrt{\sqrt{7}-\sqrt{3}}< \sqrt{\sqrt{7}+\sqrt{3}}\Rightarrow\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}< 0\)
Đặt \(x=\frac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{\sqrt{7}-2}}< 0\)
\(\Rightarrow x^2=\frac{\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}}{\sqrt{7}-2}\)
\(\Rightarrow x^2=\frac{2\sqrt{7}-2\sqrt{4}}{\sqrt{7}-2}=\frac{2\sqrt{7}-4}{\sqrt{7}-2}=\frac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
\(\Rightarrow x=-\sqrt{2}\) (do \(x< 0\))
\(\sqrt{6+2\sqrt{2}\cdot\sqrt{3-\sqrt{4+2\sqrt{3}}}}\\ =\sqrt{2\left(3+\sqrt{2}\right)\cdot\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\\ =\sqrt{2\left(3+\sqrt{2}\right)\cdot\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\\ =\sqrt{2\left(3+\sqrt{2}\right)\cdot\sqrt{3-\sqrt{3}-1}}\\ =\sqrt{2\left(3+\sqrt{2}\right)\cdot\sqrt{2-\sqrt{3}}}\)
(đến đây bó tay :<)
\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\\ =\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\\ =\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^2}}\\ =\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\\ =\sqrt{5}-\sqrt{3-2\sqrt{5}+3}=\\ \sqrt{5}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}-\sqrt{5}+1=1\)
(cái này lỡ tính rùi :3)
Câu cuối không biết, sorry nha
