Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt{3+\sqrt{5}}.\sqrt{3-\sqrt{5}}=\sqrt{3^2-\left(\sqrt{5}\right)^2}=\sqrt{4}=2\)
\(B=\sqrt{150.27.96}=\sqrt{150}.\sqrt{27}.\sqrt{96}=5\sqrt{6}.3\sqrt{3}.4\sqrt{6}=360\sqrt{3}\)
\(C=\left(\sqrt{27}+\sqrt{48}\right)^2-\left(\sqrt{27}-\sqrt{48}\right)^2\)\(=\left[\left(\sqrt{27}+\sqrt{48}-\sqrt{27}+\sqrt{48}\right)\left(\sqrt{27}+\sqrt{48}+\sqrt{27}-\sqrt{48}\right)\right]\)
\(=2\sqrt{27}.2\sqrt{48}=2.3\sqrt{3}.2.4\sqrt{3}=144\)
\(D=\sqrt{137^2-88^2}-\sqrt{192^2-111^2}=\sqrt{\left(137+88\right)\left(137-88\right)}-\sqrt{\left(192+111\right)\left(192-111\right)}\)
\(=\sqrt{225.49}-\sqrt{303.81}=15.7-9.\sqrt{303}=9\left(\frac{35}{3}-\sqrt{303}\right)\)
\(E=\sqrt{\frac{225}{4}.\frac{81}{25}.\frac{49}{64}}=\frac{15}{2}.\frac{9}{5}.\frac{7}{8}=\frac{189}{16}\)
\(F=\sqrt{\frac{27}{25}}.\sqrt{\frac{49}{189}}.\sqrt{\frac{700}{99}}=\frac{3\sqrt{3}}{5}.\frac{7}{3\sqrt{21`}}.\frac{10\sqrt{7}}{3\sqrt{11}}=\frac{14}{3\sqrt{11}}\)
\(H=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{\frac{35}{5}}+\sqrt{\frac{21}{5}}\right]=\sqrt{105}.\left[\sqrt{\frac{15}{7}}-\sqrt{7}+\sqrt{\frac{21}{5}}\right]\)
\(=\sqrt{105}.\left[\frac{\sqrt{75}-\sqrt{49}+\sqrt{147}}{\sqrt{35}}\right]=\sqrt{3}\left(12\sqrt{3}-7\right)=36-7\sqrt{3}\)
\(K=\sqrt{64.14.21.54}-\sqrt{35.45.12}=8.\sqrt{14}.\sqrt{21}.3\sqrt{6}-\sqrt{35}.3\sqrt{5}.2\sqrt{3}\)
\(=24.\sqrt{14.21.6}-6\sqrt{35.5.3}=24.42-30\sqrt{21}=30\left(\frac{168}{5}-\sqrt{21}\right)\)
a) \(A=\left(1-\sqrt{18}+\sqrt{32}\right).\sqrt{3-2\sqrt{2}}\)
\(=\left(1-\sqrt{9.2}+\sqrt{16.2}\right).\sqrt{2-2\sqrt{2}+1}\)
\(=\left(1-\sqrt{9}.\sqrt{2}+\sqrt{16}.\sqrt{2}\right).\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(1-3\sqrt{2}+4\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
\(=\left(1+\sqrt{2}\right).\left|\sqrt{2}-1\right|\)
Vì \(\sqrt{2}>1\)\(\Rightarrow\left|\sqrt{2}-1\right|>0\)
\(\Rightarrow A=\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)=\left(\sqrt{2}\right)^2-1=2-1=1\)
b) \(B=\frac{3}{6+\sqrt{35}}-\frac{3}{6-\sqrt{35}}=\frac{3\left(6-\sqrt{35}\right)}{\left(6+\sqrt{35}\right)\left(6-\sqrt{35}\right)}-\frac{3\left(6+\sqrt{35}\right)}{\left(6-\sqrt{35}\right)\left(6+\sqrt{35}\right)}\)
\(=\frac{18-3\sqrt{35}-18-3\sqrt{35}}{36-35}=-6\sqrt{35}\)
Ta có: \(M=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{224}+\sqrt{225}}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{225}-\sqrt{224}\)
\(=-1+\sqrt{225}=-1+15=14\)
Và \(N=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{63}}\)
\(=14,47706...>14=M\)