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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
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{12}-\sqrt{\left(-3\right)^2}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{2^2\cdot3}-\sqrt{3^2}\)
\(=2-\sqrt{3}+2\sqrt{3}-3\)
\(=\sqrt{3}-1\)
b) \(\left(\sqrt{8}-3\sqrt{6}+\sqrt{2}\right)\cdot\sqrt{2}+\sqrt{108}\)
\(=\sqrt{16}-3\sqrt{12}+\sqrt{4}+\sqrt{6^2\cdot3}\)
\(=4-3\sqrt{2^2\cdot3}+2+6\sqrt{3}\)
\(=6-3\cdot2\sqrt{3}+6\sqrt{3}\)
\(=6-6\sqrt{3}+6\sqrt{3}=6\)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{12}-\sqrt{\left(-3\right)^2}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{3.4}-\sqrt{3^2}=2-\sqrt{3}+\sqrt{4}.\sqrt{3}-3\)
\(=2-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}-1\)
b) \(\left(\sqrt{8}-3\sqrt{6}+\sqrt{2}\right).\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8}.\sqrt{2}-3\sqrt{6}.\sqrt{2}+\sqrt{2}.\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8.2}-3\sqrt{6.2}+2+\sqrt{36.3}\)
\(=\sqrt{16}-3\sqrt{12}+2+\sqrt{36}.\sqrt{3}\)
\(=\sqrt{4^2}-3\sqrt{4.3}+2+6\sqrt{3}\)
\(=4-3\sqrt{4}.\sqrt{3}+2+6\sqrt{3}\)
\(=4-6\sqrt{3}+2+6\sqrt{3}=6\)
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\sqrt{16}-6+\sqrt{20}-\sqrt{5}=4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2\)
b) \(0,2\sqrt{\left(-10\right)^3.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=0,2\left|-10\right|\sqrt{3}+2\left|\sqrt{3}-\sqrt{5}\right|=0,2.10.\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c) \(\left(\dfrac{1}{2}\sqrt{\dfrac{1}{2}}-\dfrac{3}{2}\sqrt{2}+\dfrac{4}{5}\sqrt{200}\right):\dfrac{1}{8}=\left(\dfrac{1}{2}\sqrt{\dfrac{2}{4}}-\dfrac{3}{2}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\left(\dfrac{1}{4}\sqrt{2}-\dfrac{2}{3}\sqrt{2}+8\sqrt{2}\right):\dfrac{1}{8}=\dfrac{27}{4}\sqrt{2}.8=54\sqrt{2}\)
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2.\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}=2\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}\)
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
a: \(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+6\sqrt{2}\right)\)
\(=\sqrt{2}\cdot4\sqrt{2}=8\)
b: \(=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{5}+1=3-\sqrt{5}\)