\(245748546\)\(\sqrt{1121221583738922\frac{12}{21545}}\)\(=?\)
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8 tháng 8 2017
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
8 tháng 9 2017
Ta có : \(\frac{3}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{\left(\sqrt{n+4}+\sqrt{n}\right)\left(\sqrt{n+4}-\sqrt{n}\right)}\)
\(=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{n+4-n}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{4}=\frac{3}{4}\left(\sqrt{n+4}-\sqrt{n}\right)\)
Áp dụng ta được :
\(\frac{3}{\sqrt{4}+\sqrt{8}}+\frac{3}{\sqrt{8}+\sqrt{12}}+\frac{3}{\sqrt{12}+\sqrt{16}}+...+\frac{3}{\sqrt{572}+\sqrt{576}}\)
\(=\frac{3}{4}\left(\sqrt{8}-\sqrt{4}+\sqrt{12}-\sqrt{8}+\sqrt{16}-\sqrt{12}+...+\sqrt{576}-\sqrt{572}\right)\)
\(=\frac{3}{4}\left(\sqrt{576}-\sqrt{4}\right)=\frac{3}{4}\left(24-4\right)=\frac{3}{4}.20=15\)
8 tháng 1 2018
Đề ở mẫu thứ 3 là \(\sqrt{xz}+12\sqrt{z}+12\) mới đúng
Ta có: \(\sqrt{xyz}=12\)
\(\Rightarrow P=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}.\sqrt{z}}{\sqrt{xz}+\sqrt{xyz}.\sqrt{x}+\sqrt{xyz}}\)\(=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{y}+\sqrt{yz}}=1\)
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14 tháng 7 2019
\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2-2.\sqrt{2}.1+1}}{\sqrt{17-3.2.2.\sqrt{2}}}-\)\(\frac{\sqrt{2+2.\sqrt{2}.1+1}}{\sqrt{17+3.2.2.\sqrt{2}}}\)
\(=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{17-3.2.\sqrt{4}.\sqrt{2}}}\)\(-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{17+3.2.\sqrt{4}.\sqrt{2}}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{8-2.\sqrt{8}.3+9}}\)\(-\frac{\sqrt{2}+1}{\sqrt{8+2.\sqrt{8}.3+9}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{\left(\sqrt{8}-3\right)^2}}\)\(-\frac{\sqrt{2}+1}{\sqrt{\left(\sqrt{8}+3\right)^2}}\)
\(=\frac{\sqrt{2}-1}{\sqrt{8}-3}\)\(-\frac{\sqrt{2}+1}{\sqrt{8}+3}\)
\(=\frac{\left(\sqrt{2}-1\right)\left(\sqrt{8}+3\right)-\left(\sqrt{2}+1\right)\left(\sqrt{8}-3\right)}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{\sqrt{16}+3\sqrt{2}-\sqrt{8}-3-\sqrt{16}+3\sqrt{2}-\sqrt{8}+3}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{6\sqrt{2}-2\sqrt{8}}{\left(\sqrt{8}-3\right)\left(\sqrt{8}+3\right)}\)
\(=\frac{6\sqrt{2}-2.2.\sqrt{2}}{\left(2\sqrt{2}-3\right)\left(2\sqrt{2}+3\right)}\)
\(=\frac{2\sqrt{2}}{\left(8-9\right)}=\frac{2\sqrt{2}}{-1}=-2\sqrt{2}\)