Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(\sqrt{12}\)+ \(2\sqrt{27}\)+\(3\sqrt{75}\)-\(9\sqrt{48}\)
b, (\(2\sqrt{2}\) - \(\sqrt{3}\))\(^2\)
a: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-14\sqrt{3}\)
b: \(\left(2\sqrt{2}-\sqrt{3}\right)^2=11-4\sqrt{6}\)
\(A=\left(3\sqrt{2}+\sqrt{6}\right)\sqrt{\frac{9-2.3\sqrt{3}+3}{2}}=\frac{\sqrt{2}\left(3+\sqrt{3}\right)}{\sqrt{2}}.\sqrt{\left(3-\sqrt{3}\right)^2}=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)
Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
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/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)
c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)
\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)
=0
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{3.4}+2\sqrt{3.9}+3\sqrt{5.5.3}-9\sqrt{4.4.3}\)
\(=2\sqrt{3}+2.3\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
=\(\sqrt{4}.\sqrt{3}+2\sqrt{9}.\sqrt{3}+3\sqrt{25}.\sqrt{3}-9\sqrt{16}.\sqrt{3}\)
=\(2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
=\(\left(2+6+15-36\right)\sqrt{3}\)
=\(-13\sqrt{3}\)