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\(\sqrt{85}.\sqrt{125}.\sqrt{68}=\sqrt{85.125.68}=\sqrt{5.17.5.25.17.4}\)
\(=\sqrt{5^2.25.17^2.4}=\sqrt{5^2}.\sqrt{25}.\sqrt{17^2}.\sqrt{4}=5.5.17.2=850\)
Ta có: \(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{81}{25}}=\sqrt{\left(\frac{9}{5}\right)^2}=\frac{9}{5}\)
\(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{27}{5}\cdot\frac{3}{5}}\)
\(=\sqrt{\frac{81}{25}}\)
\(=\sqrt{\left(\frac{9}{5}\right)^2}\)
\(=\left|\frac{9}{5}\right|=\frac{9}{5}\)
\(a\)
\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)
\(=\)\(\sqrt{2,7.1,2}\)
\(=\)\(\sqrt{3,24}\)
\(=\)\(1,8\)
\(b\)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)
\(=\)\(\sqrt{85.125.68}\)
\(=\)\(\sqrt{722500}\)
\(=\)\(850\)
\(c\)
\(\frac{\sqrt{13,5}}{\sqrt{4,5}}\)
\(=\)\(\frac{3,67}{2,12}\)
HỌC TỐT!!!
\(\frac{\sqrt{13,5}}{\sqrt{4,5}}=\sqrt{\frac{13,5}{4,5}}=\sqrt{3}\)
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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Dạng tổng quát: Với n là các số lẻ lớn hơn hoặc bằng 3 thì \(\frac{1}{n\sqrt{n-2}+\left(n-2\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n-2\right)}\left(\sqrt{n}+\sqrt{n-2}\right)}=\frac{1}{\sqrt{n\left(n-2\right)}.\frac{2}{\sqrt{n}-\sqrt{n-2}}}=\frac{\sqrt{n}-\sqrt{n-2}}{2\sqrt{n\left(n-2\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n-2}}-\frac{1}{\sqrt{n}}\right)\)Áp dụng, ta được: \(C=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{5\sqrt{3}+3\sqrt{5}}+...+\frac{1}{121\sqrt{119}+119\sqrt{121}}=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{119}}-\frac{1}{\sqrt{121}}\right)=\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{5}{11}\)Vậy C = 5/11
Xét :\(\frac{1}{\left(a+2\right)\sqrt{a}+a\sqrt{a+2}}=\frac{1}{\sqrt{a}.\sqrt{a+2}\left(\sqrt{a+2}+\sqrt{a}\right)}=\frac{\sqrt{a+2}-\sqrt{a}}{2\sqrt{a}.\sqrt{a+2}}=\frac{1}{2\sqrt{a}}-\frac{1}{2\sqrt{a+2}}\)
Xét:
\(C=\frac{1}{3\sqrt{1}+1\sqrt{3}}+\frac{1}{5\sqrt{3}+3\sqrt{5}}+...+\frac{1}{121\sqrt{119}+119\sqrt{121}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}}-\frac{1}{2\sqrt{5}}+\frac{1}{2\sqrt{5}}-\frac{1}{2\sqrt{7}}+...+\frac{1}{2\sqrt{119}}-\frac{1}{2\sqrt{121}}\)
\(=\frac{1}{2}-\frac{1}{2\sqrt{121}}=\frac{1}{2}-\frac{1}{2.11}=\frac{5}{11}\)
Xét: \(\frac{1}{n\sqrt{n-2}+\left(n-2\right)\sqrt{n}}=\frac{1}{\left(\sqrt{n}-\sqrt{n-2}\right)\sqrt{n\left(n-2\right)}}\)
\(=\frac{\sqrt{n}+\sqrt{n-2}}{2\sqrt{n\left(n-2\right)}}=\frac{1}{2}\left(\frac{\sqrt{n}+\sqrt{n-2}}{\sqrt{n\left(n-2\right)}}\right)\)
\(=\frac{1}{2}\left(\frac{1}{\sqrt{n-2}}-\frac{1}{\sqrt{n}}\right)\)
Từ đó ta thay vào:
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{199}}-\frac{1}{\sqrt{121}}\right)\)
\(C=\frac{1}{2}\cdot\left(1-\frac{1}{11}\right)\)
\(C=\frac{1}{2}\cdot\frac{10}{11}=\frac{5}{11}\)
Vậy C = 5/11
a)\(\left(\sqrt{2019.2021}\right)^2=2019.2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2\)
=> \(\sqrt{2019.2021}< 2020\)
b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>5+2\sqrt{4}=5+2.2=9\)
=> \(\sqrt{2}+\sqrt{3}>3\)
c) \(9+4\sqrt{5}=4+4\sqrt{5}+5=\left(2+\sqrt{5}\right)^2>\left(2+\sqrt{4}\right)^2=\left(2+2\right)^2=16\)
=> \(9+4\sqrt{5}>16\)
d) \(\sqrt{11}-\sqrt{3}>\sqrt{9}-\sqrt{1}=3-1=2\)
=> \(\sqrt{11}-\sqrt{3}>2\)
\(a\)
\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)
\(=\)\(\sqrt{2,7.1,2}\)
\(=\)\(\sqrt{3,24}\)
\(=\)\(1,8\)
\(b\)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)
\(=\)\(\sqrt{85.125.68}\)
\(=\)\(\sqrt{722500}\)
\(=\)\(850\)
học tốt!!!