Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đề thiếu nha:
\(E=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)(vì \(\sqrt{3}>1\))
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)
\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\) ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))
\(=\sqrt{\left(2+\sqrt{7}+2-\sqrt{7}\right)\left(2+\sqrt{7}-2+\sqrt{7}\right)}\)
\(=\sqrt{4\cdot\sqrt{7}}\)
\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}=\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2=\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)
\(=\left(3\sqrt{5}-5\sqrt{2}+5\sqrt{2}+3\sqrt{5}\right)\left(3\sqrt{5}-5\sqrt{2}-5\sqrt{2}-3\sqrt{5}\right)\)
\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)
\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)
\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}\)
cuối lười tính nên thôi nhá :>
\(1.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\text{ |}3\sqrt{2}-\sqrt{3\text{ }}\text{ |}=3\sqrt{2}-\sqrt{3}\)\(2.\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{9-2.3\sqrt{5}+5}+\sqrt{9+2.3\sqrt{5}+5}=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(3+\sqrt{5}\right)^2}=\text{ |}3-\sqrt{5}\text{ |}+\text{ |}3+\sqrt{5}\text{ |}=6\)\(3.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2.\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{ |}2\sqrt{2}+\sqrt{5}\text{ |}+\text{ |}2\sqrt{2}-\sqrt{5}\text{ |}=4\sqrt{2}\)\(4.\) Tương tự nhé bạn.
Bn ơi câu 1 cái chỗ dấu bằng thứ 1 ák lm v đc ko
\(\sqrt{21-6\sqrt{6}}=\)\(2.\sqrt{18}\sqrt{3}\)
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
\(1.\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}\)
\(2.\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}\)
\(3.\sqrt{21-6\sqrt{6}}=\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}-\sqrt{3}\)
\(4.\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(5.\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{4+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(6.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)