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\(a,A=\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\)
\(=6-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=2\)
\(b,B=\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\)
\(=\sqrt{\left(15-\sqrt{2}\right)^2+\left(11+\sqrt{2}\right)^2}\)
\(=26\)
hơi tắt
a) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\\ =\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\\ =\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\\ =\sqrt{13+30\sqrt{2}+30}\\ =\sqrt{43+30\sqrt{2}}\\ =\sqrt{\left(3\sqrt{2}+5\right)^2}\\ =3\sqrt{2}+5\)
b) \(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\\ =\sqrt{225-2\cdot15\sqrt{2}+2}+\sqrt{121+2\cdot11\sqrt{2}+2}\\ =\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}\\ =15-\sqrt{2}+11+\sqrt{2}\\ =26\)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}=\sqrt{225-30\sqrt{2}+2}+\sqrt{121+22\sqrt{2}+2}=\sqrt{15^2-15.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{11^2+11.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}=15-\sqrt{2}+11+\sqrt{2}\left(do:15-\sqrt{2}>0;11+\sqrt{2}>0\right)=26\)
a) \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right).\sqrt{\dfrac{8-2\sqrt{15}}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{25.6}-\sqrt{9.10}\right).\sqrt{\dfrac{\left(\sqrt{5}\right)^2-2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2}{2}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right).\sqrt{\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{2}}\)
\(=\left(\sqrt{10}+\sqrt{6}\right).\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\sqrt{2}.\left(\sqrt{5}+\sqrt{3}\right).\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=2\)
a) Ta có: \(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)^2\cdot\left(4+\sqrt{15}\right)\)
\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
\(=32+8\sqrt{15}-8\sqrt{15}-30\)
=2
a)
=\(\sqrt{15^2-2\cdot15\cdot\sqrt{2}+2}+\sqrt{11^2+2\cdot11\cdot\sqrt{2}+2}\)
=\(\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)}^2\)
=\(15-\sqrt{2}+11+\sqrt{2}\)
=26
c)
=\(\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}\left(\sqrt{5}+2\right)\)
=\(\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}}\)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\)
=\(\sqrt{225+2.15.\sqrt{2}+2}+\sqrt{121+2.11\sqrt{2}+2}\)
=\(\sqrt{\left(15+\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}\)
=\(15+\sqrt{2}+11+\sqrt{2}\)
=\(26+2\sqrt{2}\)