\(\sqrt{2\frac{7}{9}}.\sqrt{1\frac{24}{25}}.\sqrt{\frac{36}{25}}\)
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 3 2020
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
7 tháng 1 2017
\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)
\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)
\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)
G
17 tháng 11 2014
Bạn viết sai phân số cuối cùng.
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}+1\sqrt{2}\right)\left(2\sqrt{1}-1\sqrt{2}\right)}=\frac{2\sqrt{1}-1\sqrt{2}}{\left(2\sqrt{1}\right)^2-\left(1\sqrt{2}\right)^2}=\frac{2\sqrt{1}-1\sqrt{2}}{2^21-1^22}=\frac{2\sqrt{1}-1\sqrt{2}}{1.2}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)
Tương tự:
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{3\sqrt{2}-2\sqrt{3}}{3^22-2^23}=\frac{3\sqrt{2}-2\sqrt{3}}{2.3}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
....
\(\frac{1}{25\sqrt{24}+24\sqrt{25}}=\frac{25\sqrt{24}-24\sqrt{25}}{25^224-24^225}=\frac{25\sqrt{24}-24\sqrt{25}}{25.24}=\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
Vậy \(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=\frac{1}{1}-\frac{1}{5}=\frac{4}{5}\)
8 tháng 8 2017
Bạn trục căn thức ở mẫu rồi trừ đi là xong nhé,vì khi trục căn thức thì ở A mẫu chung là 1,ở B mẫu chung là 2.
23 tháng 8 2021
a, ĐK :a >= 3
\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)
\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)
b, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
23 tháng 8 2021
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
\(=\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{6}{5}==\frac{5\cdot7\cdot6}{3\cdot5\cdot5}=\frac{14}{5}\)
\(\sqrt{2\frac{7}{9}\cdot}\sqrt{1\frac{24}{25}}\cdot\sqrt{\frac{36}{25}}\)
\(\Leftrightarrow\sqrt{\frac{25}{9}}\cdot\sqrt{\frac{49}{25}}\cdot\sqrt{\frac{36}{25}}\)
\(\Leftrightarrow\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{6}{5}\)\(=\frac{5\cdot7\cdot6}{3\cdot5\cdot5}=\frac{210}{75}=\frac{14}{5}\)