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\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)=\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)\)\(=\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=-1\)
\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right).\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2}.\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{2+\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
\(=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
\(=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
\(=-2\)
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
C = ! x-3 ! - !x-2!
C = x - 3 - x + 2
C = -1
Dấu ! là trị tuyệt đối nha bạn