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\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)
\(=\left(\left(-\sqrt{7}\right)+\left(-\sqrt{5}\right)\right)\cdot\frac{\sqrt{7}-\sqrt{7}}{1}\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\cdot\frac{\sqrt{7}-\sqrt{5}}{1}\)
\(=\frac{-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{1}\)
\(=\frac{-\left(7-5\right)}{1}=-2\)
\(a=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}=\sqrt{3,6\left(10\right)}=\sqrt{36}=6\)
a) \(\sqrt{6,8^2-3,2^2}=\sqrt{\left(6,8-3,2\right)\left(6,8+3,2\right)}\)
=\(\sqrt{3,6.10}=\sqrt{36}=6\)
b)\(\sqrt{21,8^2-18,2^2}=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\)
=\(\sqrt{3,6.40}=\sqrt{144}=12\)
c)\(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)
=\(\sqrt{91.144-1440}=\sqrt{144.81}=\sqrt{144}.\sqrt{81}=108\)
d)\(\sqrt{146,5^2-109,5^2+27.256}\)=\(\sqrt{\left(146,5-109,5\right)\left(146,5+109,5\right)+27.256}\)
=\(\sqrt{37.256+\sqrt{27.256}}=\sqrt{64.256}=\sqrt{64}.\sqrt{256}=128\)
1,
\(a,=\sqrt{\left(21,8-18,2\right)\left(21,8+18,2\right)}\\ =\sqrt{3,6\cdot40}\\ =\sqrt{36\cdot4}\\ =\sqrt{36}\cdot\sqrt{4}\\ =6\cdot4\\ =24\)
\(b,=10\cdot\sqrt{\left(6,5-1,6\right)\left(6,5+1,6\right)}\\ =10\cdot\sqrt{4,9\cdot8,1}\\ =10\cdot\sqrt{49\cdot0,81}\\ =10\cdot\sqrt{49}\cdot\sqrt{0,81}\\ =10\cdot7\cdot0,9\\ =63\)
2,
\(A=7+4\sqrt{3}+\sqrt{3}-1\\ =6+5\sqrt{3}\\ B=7+2\sqrt{10}-\left(11+2\sqrt{10}\right)\\ =7+2\sqrt{10}-11-2\sqrt{10}\\ =-4\)
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: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}+\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=\frac{5+2\sqrt{6}+\left(5-2\sqrt{6}\right)}{3-2}=10\)
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
a;b bạn tự làm nhé, hđt số 3 là ra
c, \(\sqrt{7-2\sqrt{10}}-\sqrt{5-2\sqrt{6}}-\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{5}+\sqrt{3}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{3}+\sqrt{2}-\sqrt{5}+\sqrt{3}=0\)