Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
A=\(\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}\)
=\(\sqrt{9+6\sqrt{3}+3}+\sqrt{9-6\sqrt{3+3}}\)
=\(\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{3^2-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\)
=\(\sqrt{\left(3+\sqrt{3}\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\)
=\(3+\sqrt{3}+3-\sqrt{3}=6\)
Vậy A =6
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
\(\sqrt{12-6\sqrt{3}}=\sqrt{9-6\sqrt{3}+3}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)
\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)
\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)
\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)
\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)
\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)
Cho sửa phần mẫu số của câu trên thành \(\sqrt{6}+\sqrt{2}\)
\(\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-|2\sqrt{3}+1|}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4+2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+|\sqrt{3}-1|}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
\(\sqrt{29-4\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}=\sqrt{\left(2\sqrt{7}-1\right)^2}=\left|2\sqrt{7}-1\right|\)
\(=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}\right)^2+2.3\sqrt{2}.1+1^2}=\sqrt{\left(3\sqrt{2}+1\right)^2}=\left|3\sqrt{2}+1\right|\)
\(=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.1+1^2}=\sqrt{\left(3\sqrt{3}-1\right)^2}=\left|3\sqrt{3}-1\right|\)
\(=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}\right)^2-2.3\sqrt{5}.1+1^2}=\sqrt{\left(3\sqrt{5}-1\right)^2}=\left|3\sqrt{5}-1\right|\)
\(=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=\sqrt{\left(4\sqrt{3}\right)^2+2.4\sqrt{3}.1+1^2}=\sqrt{\left(4\sqrt{3}+1\right)^2}=\left|4\sqrt{3}+1\right|\)
\(=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.2+2^2}=\sqrt{\left(2\sqrt{7}-2\right)^2}=\left|2\sqrt{7}-2\right|\)
\(=2\sqrt{7}-2\)
\(\sqrt{29-4\sqrt{7}}=2\sqrt{7}-1\)
\(\sqrt{19+6\sqrt{2}}=3\sqrt{2}+1\)
\(\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
\(\sqrt{46-6\sqrt{5}}=3\sqrt{5}-1\)
\(\sqrt{49+8\sqrt{3}}=4\sqrt{3}+1\)
\(\sqrt{32-8\sqrt{7}}=2\sqrt{7}-2\)
\(\sqrt{28-6\sqrt{3}}-\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}-\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=3\sqrt{3}-1-3-\sqrt{3}=2\sqrt{3}-4\)
học tốt ~