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a. \(=\sqrt{2}.\left(\sqrt{7}+\sqrt{8}\right)\sqrt{5-\sqrt{3}\sqrt{7}}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{3-2\sqrt{3}.\sqrt{7}+7}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{8}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
Rồi nhân ra. bạn làm tiếp nhé. Tuy nhiên minh nghĩ bạn bị nhầm đề. là \(\sqrt{6}\) chứ không phải căn 16
b. \(=\frac{5\left(\sqrt{21}+1\right)}{21-16}+\frac{\sqrt{3}.\sqrt{7}\left(\sqrt{3}-\sqrt{7}\right)}{-\left(\sqrt{3}-\sqrt{7}\right)}\)
\(=\sqrt{21}+4-\sqrt{21}=4\)
\(\text{a)}\)\(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-\sqrt{250}\)
\(\Leftrightarrow5\sqrt{10}+10-5\sqrt{10}\)
\(\Leftrightarrow10\)
\(\text{b)}\)\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-2\sqrt{21}-7+2\sqrt{21}\)
\(\Leftrightarrow4\sqrt{21}-7\)
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)
= \(6-\sqrt{15}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)
= \(7\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)
a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15
b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10
c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7
d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22
Ta có: \(A=\left(5+\frac{7-\sqrt{21}}{\sqrt{7}-\sqrt{3}}\right)\sqrt{16-5\sqrt{7}}\)
\(=\left(5+\frac{\sqrt{7}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{7}-\sqrt{3}}\right)\sqrt{16-5\sqrt{7}}\)
\(=\left(5+\sqrt{7}\right)\sqrt{16-5\sqrt{7}}\)
\(\Rightarrow\sqrt{2}A=\left(5+\sqrt{7}\right)\sqrt{32-10\sqrt{7}}\)
\(=\left(5+\sqrt{7}\right)\sqrt{\left(5-\sqrt{7}\right)^2}=\left(5+\sqrt{7}\right)\left(5-\sqrt{7}\right)\)
\(=25-7=18\)
Vậy \(A=\frac{18}{\sqrt{2}}=9\sqrt{2}\)