A) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
b) \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
c) \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
d)...
Đọc tiếp
A) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
b) \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
c) \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
d) \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
e)\(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
g)\(\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)
\(=\sqrt{3}\left(2+4-5-9\right)\)
\(=-8\sqrt{3}\)
b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)
\(=7-5=2\)
c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\cdot\left|\sqrt{3}-1\right|\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)
\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)
\(=-11\sqrt{2}\)
e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
\(=\left(\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=3-5=-2\)
g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)
\(=\left(\sqrt{3}-1\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\cdot\left|\sqrt{3}+1\right|\)
\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))
\(=3-1=2\)