\(\sqrt[2]{\frac{1}{4}}\)-\(\sqrt{25}\)+\(\sqrt{\frac{4}{9}}\)
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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 \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
2 tháng 12 2017
\(\sqrt{\frac{25}{4}}+\left(\sqrt{\frac{1}{2}}\right)^2:\left(\frac{-\sqrt{9}}{4}\right).\sqrt{\frac{16}{81}}-4^2-\left(-2\right)^3\)
\(=\frac{5}{2}+\frac{1}{2}:\frac{-3}{4}.\frac{4}{9}-16+8\)
\(=\frac{5}{2}-\frac{8}{27}-8\)
\(=\frac{-313}{54}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2019
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\Leftrightarrow2\sqrt{2x-1}-\frac{5}{2}\sqrt{2x-1}+\frac{2}{3}\sqrt{2x-1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{6}\sqrt{2x-1}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{2x-1}=3\)
\(\Leftrightarrow2x-1=9\Rightarrow x=5\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 3 2020
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 5 2019
Đặt \(x=\sqrt[4]{5}\Rightarrow x^4=5\Rightarrow x^4-5=0\)
\(A=\frac{2}{\sqrt{4-3x+2x^2-x^3}}=\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}\)
\(=\frac{2\left(x+1\right)}{\sqrt{4+5x-x^5}}=\frac{2\left(x+1\right)}{\sqrt{4+x\left(5-x^4\right)}}=x+1=\sqrt[4]{5}+1\)
\(B=\left(\frac{-\sqrt[4]{2}\left(1-\sqrt[4]{2}\right)}{1-\sqrt[4]{2}}+\frac{1+\sqrt{2}}{\sqrt[4]{2}}\right)^2-\frac{\sqrt{1+\sqrt{2}+\frac{1}{2}}}{1+\sqrt{2}}\)
\(=\left(-\sqrt[4]{2}+\frac{1}{\sqrt[4]{2}}+\sqrt[4]{2}\right)^2-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)
\(=\frac{1}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\)
\(\sqrt[2]{\frac{1}{4}}-\sqrt{25}+\sqrt{\frac{4}{9}}=\frac{1}{2}-5+\frac{2}{3}=\)\(\frac{-23}{6}\)