\(\sqrt{\sqrt{2}-\sqrt{30+4\sqrt{6}-8\sqrt{2}-16\sqrt{3}}}\) Câu này giải sao vậy chỉ mình với
có ai biết giải ko giải hộ mình mấy bài này với ( giải chi tiết hộ mình nhé)
1, \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
2, \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
3, \(\sqrt{4+\sqrt{5\sqrt{3+}5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
4, \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
5, \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
6, \(\sqrt{4+\sqrt{8}.\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
7, \(\sqrt{8\sqrt{3}-2\sqrt{25\sqrt{12}+4\sqrt{192}}}\)
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)
\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)
\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)
\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)
5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}-\sqrt{2}\right)}{5\left(\sqrt{3}-\sqrt{2}\right)}=1\)
có ai biết giải bài này không giúp mình với mình đang cần gấp, xin cảm ơn
Bài 20: rút gọn
1, \(\sqrt{9-4\sqrt{5}}.\sqrt{9+4\sqrt{5}}\)
2, \(\left(2\sqrt{2}-6\right).\sqrt{11+6\sqrt{2}}\)
3, \(\sqrt{2}.\sqrt{2-\sqrt{3}}\left(\sqrt{3}+1\right)\)
4, \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
5, \(\sqrt{27+10\sqrt{2}}:\dfrac{1}{\sqrt{\left(\sqrt{2}-5\right)^2}}\)
Bài 21: rút gọn
1, \(5\sqrt{\dfrac{1}{5}}\) 2, \(\dfrac{12}{5}\sqrt{\dfrac{5}{4}}\)
3, \(\dfrac{30}{5\sqrt{6}}\) 4, \(\dfrac{20}{2\sqrt{5}}\)
5, \(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
a)\(\sqrt{11-6\sqrt{2}}-\sqrt{27+10\sqrt{2}}\)
b) \(\sqrt{13-4\sqrt{3}}-\sqrt{16-8\sqrt{3}}\)
giải giúp mình với
a) \(\sqrt{11-6\sqrt{2}}-\sqrt{27+10\sqrt{2}}\)
\(=\sqrt{9-6\sqrt{2}+2}-\sqrt{25+10\sqrt{2}+2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(5+\sqrt{2}\right)^2}\)
\(=\left|3-\sqrt{2}\right|-\left|5+\sqrt{2}\right|\)
\(=3-\sqrt{2}-5-\sqrt{2}=-2-2\sqrt{2}\)
b) \(\sqrt{13-4\sqrt{3}}-\sqrt{16-8\sqrt{3}}\)
\(=\sqrt{12-4\sqrt{3}+1}-\sqrt{12-8\sqrt{3}+4}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}-\sqrt{\left(2\sqrt{3}-2\right)^2}\)
\(=\left|2\sqrt{3}-1\right|-\left|2\sqrt{3}-2\right|\)
\(=2\sqrt{3}-1-2\sqrt{3}+2\)
\(=1\)
giải hộ mik
a)\(\sqrt{11+6\sqrt{2}}-\left(3+\sqrt{2}\right)\)
b)\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\sqrt{11+6\sqrt{2}}-\left(3+\sqrt{2}\right)\)
\(=3+\sqrt{2}-3-\sqrt{2}\)
=0
b) Ta có: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}+\sqrt{8}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2}\)
\(=\sqrt{2}+1\)
Giúp mình tính bài này với ạ =)) cần gấp.
a) \(\left(2+\sqrt{4+\sqrt{6-2\sqrt{5}}}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
b) \(\sqrt{6-2\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}\)
c) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
d) \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}-3\sqrt{2}\)
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/ \(=\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)
\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)
\(=\sqrt{13+6\left(\sqrt{3}+1\right)}\)
\(=\sqrt{19+6\sqrt{2}}\)
\(=3\sqrt{2}+1\)
Ai giải cho mình phép này với, mình không biết bắt đầu từ đâu:
\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(\sqrt{2}\sqrt{4}\sqrt{3}+2.5\sqrt{4}\sqrt{3}-4\sqrt{16}\sqrt{4}\sqrt{3}=2\sqrt{6}+10\sqrt{12}-16\sqrt{12}=2\sqrt{6}-6\sqrt{12}=2\sqrt{6}-12\sqrt{3}\)
có ai biết giải bài toán này k giúp mình với ?
1,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}}\)
2,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
3,\(\dfrac{3}{\sqrt{6}-\sqrt{3}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
4,\(\left(\sqrt{\dfrac{2}{3}-\sqrt{\dfrac{3}{2}}+\dfrac{5}{\sqrt{6}}}\right):\dfrac{6-\sqrt{6}}{1-\sqrt{6}}\)
5,\(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\times\left(\sqrt{3}+\sqrt{2}\right)\)
6,\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
7, \(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
8,\(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
9,\(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
10,\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\)
11,\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
12,\(\dfrac{\sqrt{3}+2\sqrt{2}+\sqrt{3}-2\sqrt{2}}{\sqrt{3}+2\sqrt{2}-\sqrt{3}-2\sqrt{2}}\)
1. Sửa đề:
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)
2.
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)
3.
\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\)
\(=\frac{3(\sqrt{6}+\sqrt{3})}{3}+\frac{4(\sqrt{7}-\sqrt{3})}{4}=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
giải giúp mình câu này với
\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\) = \(\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{-\dfrac{4}{1+\sqrt{5}}}\) = \(2\sqrt{5}-2\left(1+\sqrt{5}\right)\) = -2.
bài 1: thực hiện phép tính
1) 2\(\sqrt{5}\)\(-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
3) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
4) \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}-\sqrt{27}}{\sqrt{30}-\sqrt{162}}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
6) \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}-\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}}\)
GIÚP MÌNH VỚI M.N~~~!!!!!!!!! MÌNH ĐANG CẦN GẤP LẮM (T_T)
1) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)
\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)
\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)
\(=4\sqrt{5}\)
2) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-\sqrt{6^2.6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-6\sqrt{6}+3^2}+\sqrt{\left(2\sqrt{6}\right)^2-12\sqrt{6}+3^2}\)
\(=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|\sqrt{6}-3\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vi \(\sqrt{6}-3< 0\))
\(=\sqrt{6}\)
5) \(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
\(=2\frac{4}{\sqrt{3}}-3.\frac{1}{3}-6\sqrt{\frac{2^2}{3.5^2}}\)
\(=\frac{8\sqrt{3}}{3}-1-6.\frac{2}{5}.\sqrt{\frac{1}{3}}\)
\(=8\frac{\sqrt{3}}{3}-1-\frac{12}{5}.\frac{\sqrt{3}}{3}\)
\(=\frac{28}{5}.\frac{\sqrt{3}}{3}-1\)
Báo cáo sai phạm
1) 2√5−√125−√80+√605
=2√5−√52.5−√42.5+√112.5
=2√5−5√5−4√5+11√5
=4√5
2) √15−√216+√33−12√6
=√15−√62.6+√33−12√6
=√15−6√6+√33−12√6
=√(√6)2−6√6+32+√(2√6)2−12√6+32
=√(√6−3)2+√(2√6−3)2
=|√6−3|+|2√6−3|
=3−√6+2√6−3 ( vi √6−3<0)
=√6
5) 2√163 −3√127 −6√475
=24√3 −3.13 −6√223.52
=8√33 −1−6.25 .√13
=8√33 −1−125 .√33
=285 .√33 −1