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Ta có : \(A=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{2^2-2.2\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
Ta có : \(B=\sqrt{3+\sqrt{8}+\sqrt{3-\sqrt{8}}}\)
\(=\sqrt{3+\sqrt{8}+\sqrt{2-2\sqrt{2}+1}}\)\(=\sqrt{3+\sqrt{8}+\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}+\sqrt{2}-1}\) \(=\sqrt{2+3\sqrt{2}}\)
b,\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\) \(=\sqrt{8\sqrt{3}}-2\sqrt{50\sqrt{3}}+4\sqrt{8\sqrt{3}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}\)
\(=0\)
d,\(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{2}(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}})\)
\(\sqrt2A=\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
\(\sqrt2A=\sqrt{(\sqrt5-1)^2}\) \(+\sqrt{(\sqrt5+1)^2}\) \(=\sqrt5-1 +\sqrt5+1=2\sqrt5\)
\(\Rightarrow A=\dfrac{2\sqrt5}{\sqrt2}\) \(=\sqrt{10}\)
a. \(\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\frac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{3\sqrt{5}-3+5-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=\frac{2\left(\sqrt{5}+1\right)}{2\left(\sqrt{5}+1\right)}=1\)
a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=1\)
b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{2}-2\cdot5\sqrt{2}+4\cdot8\sqrt{2}\right)\)
\(=\sqrt{\sqrt{3}}\cdot24\sqrt{2}\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
Mik sẽ viết lại đề bài.Bạn cs thể giải đầy đủ cho mik giùm nhen ko cần ngắn cứ dài . Cảm ơn
A=\(\sqrt{7}-4\sqrt{3}+\sqrt{4}-2\sqrt{3}\)
B=\(\left(2+\frac{5-\sqrt{5}}{\sqrt{5}-1}\right)\) \(\left(2-\frac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
C=\(\left(\sqrt{3}+1\right)\) \(\frac{\sqrt{14}-6\sqrt{3}}{5+\sqrt{3}}\)
nguyen thao:
Câu A: vẫn giống ban đầu mà bạn? Mình nghĩ bạn vẫn viết sai đề. Đề đúng là \(A=\sqrt{7-4\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(B=\left[2+\frac{\sqrt{5}(\sqrt{5}-1)}{\sqrt{5}-1}\right]\left[2-\frac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}+1)}\right]\)
\(=(2+\sqrt{5})(2-\sqrt{5})=2^2-(\sqrt{5})^2=4-5=-1\)
$C=\frac{(\sqrt{3}+1)(\sqrt{14}-6\sqrt{3})}{5+\sqrt{3}}$
$=\frac{-18-6\sqrt{3}+\sqrt{14}+\sqrt{42}}{5+\sqrt{3}}$ vẫn xấu lắm bạn ạ :''>
c/ = \(\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{43+30\sqrt{2}}\)
\(=\sqrt{25+2.3.5.\sqrt{2}+18}\)
\(=5+3\sqrt{2}\)
d/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
Bài làm:
a) \(A=\sqrt{4}-2\sqrt{3}+\sqrt{7}-4\sqrt{3}\)
\(A=2+\sqrt{7}-6\sqrt{3}\)
b) \(B=\sqrt{3}+\sqrt{8}+\sqrt{3}-\sqrt{8}\)
\(B=2\sqrt{3}\)