Câu hỏi của Linh

Question

d)$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}$

e)$\frac{2}{\sqrt{3}-1}$-$\frac{3-2\sqrt{3}}{2-\sqrt{3}}$

2 câu cần gấp ạ

Akai Haruma · Accepted Answer

Lời giải:
d)

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}$

$=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}$

$=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}$

$=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}$

$=\sqrt{3}+1$

e)

$\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}$

$=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1$Câu trả lời: Lời giải:
d)

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}$

$=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}$

$=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}$

$=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}$

$=\sqrt{3}+1$

e)

$\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}$

$=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1$

Yuzu · Answer

d)

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\
=\sqrt{6-2\sqrt{3}-2}\
=\sqrt{4-2\sqrt{3}}\
=\sqrt{3-2\sqrt{3}+1}\
=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1$

e)

$\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\
=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\
=\frac{2}{\sqrt{3}-1}+\sqrt{3}\
=\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\
=\frac{5-\sqrt{3}}{\sqrt{3}-1}\
=\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\
-\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4$

(bạn nhớ ktr đã nha)Câu trả lời: d)

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\
=\sqrt{6-2\sqrt{3}-2}\
=\sqrt{4-2\sqrt{3}}\
=\sqrt{3-2\sqrt{3}+1}\
=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1$

e)

$\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\
=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\
=\frac{2}{\sqrt{3}-1}+\sqrt{3}\
=\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\
=\frac{5-\sqrt{3}}{\sqrt{3}-1}\
=\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\
-\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4$

(bạn nhớ ktr đã nha)

Y · Answer

d) $=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}$

$=\sqrt{6+2\sqrt{5-\sqrt{12}-1}}$ $=\sqrt{6+2\sqrt{4-\sqrt{12}}}$

$=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}$

$=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}$

$=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1$

e) $=\frac{2\left(2-\sqrt{3}\right)-\left(\sqrt{3}-1\right)\left(3-2\sqrt{3}\right)}{\left(\sqrt{3}-1\right)\left(2-\sqrt{3}\right)}$ $=\frac{4-2\sqrt{3}-3\sqrt{3}+3+6-2\sqrt{3}}{2\sqrt{3}-2+\sqrt{3}-3}$

$=\frac{13-7\sqrt{3}}{3\sqrt{3}-5}=\frac{3\sqrt{3}+18-5-10\sqrt{3}}{3\sqrt{3}-5}$

$=\frac{3\sqrt{3}-5+2\sqrt{3}\left(3\sqrt{3}-5\right)}{3\sqrt{3}-5}=\frac{\left(3\sqrt{3}-5\right)\left(1+2\sqrt{3}\right)}{3\sqrt{3}-5}$ $=1+2\sqrt{3}$Câu trả lời: d) $=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}$