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\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)
Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\frac{7-3}{2}=2\)
Vậy \(P=2\)
Câu A=4
Cách giải:
\(\left(5\sqrt{3}+2\sqrt{12}-\sqrt{75}\right):\sqrt{3}\)
\(=\left(5\sqrt{3}+2\sqrt{4\cdot3}-\sqrt{25\cdot3}\right)\)\(:\sqrt{3}\)
\(=\left(5\sqrt{3}+4\sqrt{3}-5\sqrt{3}\right)\)\(:\sqrt{3}\)
\(a.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}\)
\(=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=\sqrt{\left(16-15\right)}=\sqrt{1}=1\)
\(b.\sqrt{7-\sqrt{47}}.\sqrt{14+2\sqrt{47}}\)
\(=\sqrt{7-\sqrt{47}}.\sqrt{2\left(7-\sqrt{47}\right)}\)
\(=\sqrt{2\left(7-\sqrt{47}\right)\left(7+\sqrt{47}\right)}=\sqrt{2\left(49-47\right)}=\sqrt{2^2}=\sqrt{4}=2\)
\(c.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=\sqrt{16-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)
\(=\sqrt{16-10-2\sqrt{5}}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)
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+) Ta có: \(2\sqrt{75}-4\sqrt{27}+3\sqrt{12}\)
\(=2\sqrt{25}.\sqrt{3}-4\sqrt{9}.\sqrt{3}+3\sqrt{4}.\sqrt{3}\)
\(=10.\sqrt{3}-12.\sqrt{3}+6.\sqrt{3}\)
\(=4\sqrt{3}\approx6,9282\)
+) Ta có:\(\sqrt{x+6\sqrt{x-9}}\)
\(=\sqrt{x-9+6\sqrt{x-9}+9}\)
\(=\sqrt{\left(\sqrt{x-9}-3\right)^2}\)
\(=\left|\sqrt{x-9}-3\right|\)
\(\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{2-\sqrt{3}}=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{2+\sqrt{3}}{4-3}=\sqrt{5}-\sqrt{3}+2+\sqrt{3}=\sqrt{5}+2\)
\(C=\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{\frac{4-2\sqrt{3}}{2}}.\left[\sqrt{2}.\left(\sqrt{3}+\sqrt{1}\right)\right]\)
\(=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)
\(=\frac{\sqrt{3}-1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
\(D=\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)
\(=\frac{\left(8+2\sqrt{2}\right).\left(3+\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}.\left(2+3\sqrt{2}\right)}{2}+\frac{\sqrt{2}.\left(1+\sqrt{2}\right)}{1-2}\)
\(=\frac{24+14\sqrt{2}+4}{7}-\frac{2\sqrt{2}+6}{2}-\frac{\sqrt{2}+2}{1}\)
\(=\frac{28+14\sqrt{2}}{7}-\sqrt{2}-3-\sqrt{2}-2\)
\(=4+2\sqrt{2}-2\sqrt{2}-5\)
\(=-1\)
\(\sqrt{8-2\sqrt{15}}+\sqrt{48+6\sqrt{15}}\\ =\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{3}+3}\\ =\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}+3\sqrt{5}+\sqrt{3}=4\sqrt{5}\)
\(\sqrt{8-\sqrt{60}}-\sqrt{23-\sqrt{240}}\\ =\sqrt{8-\sqrt{4\cdot15}}-\sqrt{23-\sqrt{4\cdot60}}\\ =\sqrt{8-2\sqrt{15}}-\sqrt{23-2\sqrt{60}}\\ =\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{20-2\cdot\sqrt{20}\cdot\sqrt{3}+3}\\ =\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-\sqrt{3}\right)^2}\\ =\sqrt{5}-\sqrt{3}-\sqrt{20}+\sqrt{3}\\ =\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)