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\(F=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{2012}-\sqrt{2011}}{\left(\sqrt{2012}+\sqrt{2011}\right)\left(\sqrt{2012}-\sqrt{2011}\right)}\)
\(F=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{2012}-\sqrt{2011}}{2012-2011}\)
\(F=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2012}-\sqrt{2011}\)
\(F=\sqrt{2012}-\sqrt{1}\)
\(F=\sqrt{2012}-1\)
\(\sqrt{\left(5+2\sqrt{6}\right)}+\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{\left(2+2.\sqrt{2}.\sqrt{3}+3\right)}+\sqrt{3-2\sqrt{3}.\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{3}+\sqrt{5}\)
\(=\sqrt{2}+2\sqrt{3}+\sqrt{5}\)
Ta có: \(\sqrt{5}< \sqrt{16}\left(=4\right)\)
\(\Rightarrow\sqrt{5}-4< 0\)
\(mà2\sqrt{3}>0\)
\(\Rightarrow2\sqrt{3}>\sqrt{5}-4\)
Ta có: \(\sqrt{5}< \sqrt{12}\)và \(-4< 0\)
\(\Rightarrow\sqrt{5}-4< \sqrt{12}+0=\sqrt{12}\)
mà \(\sqrt{12}=\sqrt{4.3}=\sqrt{4}.\sqrt{3}=2\sqrt{3}\)
\(\Rightarrow2\sqrt{3}=\sqrt{5}-4\)
mãi mãi mới có 1 bài đây nè
Câu hỏi của Nguyễn Thùy Linh - Toán lớp 9 - Học toán với OnlineMath
vào thóng kê
k 3 phát như đã hứa nhé
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\(\sqrt{\frac{1}{2}}+\sqrt{4,5}-\sqrt{12,5}-0,5\sqrt{200}+\sqrt{242}+6\sqrt{1\frac{1}{8}}-\sqrt{24,5}\)
\(=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}-\frac{5\sqrt{2}}{2}-5\sqrt{2}+11\sqrt{2}+\frac{9\sqrt{2}}{2}-\frac{7\sqrt{2}}{2}\)
\(=\frac{\sqrt{2}}{2}+6\sqrt{2}\)
\(=\frac{13\sqrt{2}}{2}\)
\(a,\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}\)
\(=|\sqrt{2}-3|.\sqrt{9+6\sqrt{2}+2}\)
\(=(3-\sqrt{2}).\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
\(=9-2=7\)
\(b,\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\frac{1}{3-\sqrt{3}}}\)
\(=\left(3-\sqrt{3}\right).\frac{\sqrt{1}}{\sqrt{3-\sqrt{3}}}\)
\(=\frac{3-\sqrt{3}}{\sqrt{3-\sqrt{3}}}\)
\(=\sqrt{3-\sqrt{3}}\)
\(c,-\frac{2}{3}\sqrt{\frac{\left(a-b\right)^3.b^5}{c}}.\frac{9}{4}\sqrt{\frac{c^3}{2\left(a-b\right)}}.\sqrt{98b}\)
\(=-\frac{2}{3}.\frac{\sqrt{\left(a-b\right)^3.b^5}}{\sqrt{c}}.\frac{9}{4}.\frac{\sqrt{c^3}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{\left(a-b\right)b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{9}{4}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.7\sqrt{2b}\)
\(=-\frac{2}{3}.\frac{9}{4}.7.\frac{\left(a-b\right).b^2\sqrt{\left(a-b\right)b}}{\sqrt{c}}.\frac{c\sqrt{c}}{\sqrt{2\left(a-b\right)}}.\sqrt{2b}\)
\(=-\frac{21}{2}.\left(a-b\right).b^2\sqrt{b}.c.\sqrt{b}\)
\(=\frac{-21}{2}.\left(a-b\right).b^3.c\)
\(d,\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\frac{1}{2}.2\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+5\sqrt{2}-\sqrt{2}\right).2\sqrt{6}\)
\(=\left(\sqrt{6}-3\sqrt{3}+4\sqrt{2}\right).2\sqrt{6}\)
\(=2.6-18\sqrt{2}+16\sqrt{3}\)
\(=12-18\sqrt{2}+16\sqrt{3}\)