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a: \(=\sqrt{7+\sqrt{45}}-\sqrt{7-\sqrt{45}}\)
\(=\dfrac{\sqrt{14+2\sqrt{45}}-\sqrt{14-2\sqrt{45}}}{\sqrt{2}}\)
\(=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
b: \(=2\cdot\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\cdot4\sqrt{5}-2\sqrt{5}-3\cdot2\sqrt{5}\right)\)
\(=\sqrt{\sqrt{3}}\cdot0=0\)
c: \(=\left(2-\sqrt{3}-6+2\sqrt{3}+8+2\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
\(=\left(4+3\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
=16-27=-11
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
a) \(\sqrt{\left|1-\sqrt{3}\right|^2}\cdot\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\sqrt{3+2\sqrt{3}+1}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}\right)^2-1^2=3-1=2\)
b) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot4\sqrt{3}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
a) √∣∣1−√3∣∣2⋅√4+2√3|1−3|2⋅4+23
=√(√3−1)2⋅√3+2√3+1=(3−1)2⋅3+23+1
=(√3−1)√(√3+1)2=(3−1)(3+1)2
=(√3−1)(√3+1)=(3−1)(3+1)
=(√3)2−12=3−1=2=(3)2−12=3−1=2
b) 2√40√12−2√√75−3√5√4824012−275−3548
=2√40⋅2√3−2√5√3−3√5⋅4√3=240⋅23−253−35⋅43
=2√80√3−2√5√3−3√20√3=2803−253−3203
=8√5√3−2√5√3−6√5√3=0