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I: Rút gọn
\(A=\sqrt{7-4\sqrt{3}}\\ =\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}\\ =2-\sqrt{3}\)
\(B=\sqrt{19-8\sqrt{3}}\\ =\sqrt{16-2\cdot4\cdot\sqrt{3}+3}\\ =\sqrt{\left(4-\sqrt{3}\right)^2}\\ =4-\sqrt{3}\)
\(C=\sqrt{21-4\sqrt{5}}\\ =\sqrt{20-2\cdot2\sqrt{5}+1}\\ =\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot1+1}\\ =\sqrt{\left(2\sqrt{5}-1\right)^2}\\ =2\sqrt{5}-1\)
Câu D mình làm chưa ra, sorry :<
a,\(\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}\)
\(=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)
\(=3-1\)
\(=2\)
b, \(\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)\)
\(=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\sqrt{2}\left(5-3\right)\)
\(=2\sqrt{2}\)
1: \(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)\)
\(=2\left(2+\sqrt{3}\right)=4+2\sqrt{3}\)
2: \(=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
3: \(=\left(\sqrt{7}-\sqrt{2}+\sqrt{7}+\sqrt{2}\right)^2=\left(2\sqrt{7}\right)^2=28\)
\(a,\left(\sqrt{50}+\sqrt{48}-\sqrt{72}\right)2\sqrt{3}\)
\(=\left(5\sqrt{2}+4\sqrt{3}-6\sqrt{2}\right)2\sqrt{3}\)
\(=\left(4\sqrt{3}-\sqrt{2}\right)2\sqrt{3}\)
\(=24-2\sqrt{6}\)
b) Ta có: \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(\frac{5+2\sqrt{6}+2\left(5-2\sqrt{6}\right)}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=\frac{5+2\sqrt{6}+10-4\sqrt{6}}{25-24}\cdot\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\cdot\left(15+2\sqrt{6}\right)\)
\(=15^2-\left(2\sqrt{6}\right)^2\)
\(=225-24=201\)
A=căn[(x-5)2]/x-5=|x-5|/x-5
Nếu x>=5 thì A=1
Nếu x<5 thì A=-1