Tính :
\(\sqrt{9}+9\times\sqrt{16}-\frac{15}{3}=???\)
(dễ đúng không)
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Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6\cdot\frac{\sqrt{3}}{2}+10\cdot\frac{2\sqrt{3}}{5}-15\cdot\frac{4}{\sqrt{3}}+9\cdot\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}=-7\sqrt{3}\)
Trả lời:
\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)
\(=6.\frac{\sqrt{3}}{\sqrt{4}}+10.\frac{\sqrt{12}}{\sqrt{25}}-15.\frac{\sqrt{16}}{\sqrt{3}}+9.\frac{\sqrt{4}}{\sqrt{3}}\)
\(=6.\frac{\sqrt{3}}{2}+10.\frac{\sqrt{2^2.3}}{5}-15.\frac{4}{\sqrt{3}}+9.\frac{2}{\sqrt{3}}\)
\(=3\sqrt{3}+10.\frac{2\sqrt{3}}{5}-15.\frac{4\sqrt{3}}{3}+9.\frac{2\sqrt{3}}{3}\)
\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}\)
\(=\left(3+4-20+6\right).\sqrt{3}=-7\sqrt{3}\)
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
\(\sqrt{9}+9\times\sqrt{16}-\frac{15}{3}=a\)
\(3+9\times4-5=a\)
\(3+36-5=a\)
\(39-5=a\)
\(34=a\)
= 3 + 9 x 4 - 15/3
Rồi tính tiếp đi :)