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câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
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: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
https://hoc24.vn/hoi-dap/question/407636.html
\(M=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+5}\)
= 9
~ ~ ~ ~ ~
\(M=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{6+2\sqrt{3}-2}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}+1\)
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 \(\sqrt{3}>1\))
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
\(D=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\Rightarrow D\sqrt{2}=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
\(\Rightarrow D=\frac{2}{\sqrt{2}}=\sqrt{2}\)
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
\(f,\sqrt{\dfrac{3-\sqrt{5}}{2-\sqrt{3}}}\\ =\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}{4-3}}\\ =\sqrt{\left(3-\sqrt{5}\right)\left(2+\sqrt{3}\right)}\\ =\sqrt{\dfrac{\left(6-2\sqrt{5}\right)\left(4+2\sqrt{3}\right)}{4}}\\ =\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{3}+1\right)}{2}\)
\(a,\sqrt{3+\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)\\ =\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{3-\sqrt{5}}.\sqrt{2}\left(\sqrt{5}+1\right)\\ =\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}.\sqrt{6-2\sqrt{5}}.\left(\sqrt{5}+1\right)\\ =\sqrt{9-5}.\sqrt{\left(\sqrt{5}-1\right)^2}.\left(\sqrt{5}+1\right)\\ =2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\\ =2.4\\ =8\)
a)
\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)
\(=2(4^2-15)=2\)
b)
\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)
\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)
c)
\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)
\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)
\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)
d)
\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)
\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)
Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)
\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)