\(\frac{\sqrt{13,5}}{\sqrt{4,5}}=\sqrt{\frac{13,5}{4,5}}=\sqrt{3}\)

\(\sqrt{85}.\sqrt{125}.\sqrt{68}=\sqrt{85.125.68}=\sqrt{5.17.5.25.17.4}\)

\(=\sqrt{5^2.25.17^2.4}=\sqrt{5^2}.\sqrt{25}.\sqrt{17^2}.\sqrt{4}=5.5.17.2=850\)

\(a\)

\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)

\(=\)\(\sqrt{2,7.1,2}\)

\(=\)\(\sqrt{3,24}\)

\(=\)\(1,8\)

\(b\)

\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)

\(=\)\(\sqrt{85.125.68}\)

\(=\)\(\sqrt{722500}\)

\(=\)\(850\)

học tốt!!!

Giả sử \(8< \sqrt{15}+\sqrt{17}\)

\(\Leftrightarrow64< 15+2\sqrt{15.17}+17\)(Bình phương hai vế)

\(\Leftrightarrow32< 2\sqrt{15.17}\)

\(\Leftrightarrow16< \sqrt{15.17}\)

\(\Leftrightarrow16< \sqrt{\left(16-1\right)\left(16+1\right)}\)

\(\Leftrightarrow\sqrt{16^2}< \sqrt{16^2-1}\)

\(\Leftrightarrow16^2< 16^2-1\)(vô lí)

Chứng minh tương tự điều giả sử \(8=\sqrt{15}+\sqrt{17}\)

Vậy \(8>\sqrt{15}+\sqrt{17}\)

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Ta có: \(\sqrt{2,7}\cdot\sqrt{1,2}\)

\(=\sqrt{2,7\cdot1,2}\)

\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)

\(=\sqrt{\frac{81}{25}}=\sqrt{\left(\frac{9}{5}\right)^2}=\frac{9}{5}\)

\(\sqrt{2,7}\cdot\sqrt{1,2}\)

\(=\sqrt{2,7\cdot1,2}\)

\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)

\(=\sqrt{\frac{27}{5}\cdot\frac{3}{5}}\)

\(=\sqrt{\frac{81}{25}}\)

\(=\sqrt{\left(\frac{9}{5}\right)^2}\)

\(=\left|\frac{9}{5}\right|=\frac{9}{5}\)

Bạn ấy sai thì bạn nhắc nhẹ thôi chứ làm gì phải ồ zê như vậy