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Bạn Duy giúp mik vs
Tính:
a) \(\sqrt{0,36}+\sqrt{0,49}\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=0,6+0,7=1,3
b)\(\sqrt{\frac{4}{9}}\)-\(\sqrt{\frac{25}{36}}\)=2/3-5/6=4/6-5/6=-1/6
a) \(\sqrt{0,36}+\sqrt{0,49}=\sqrt{\left(0,6\right)^2}+\sqrt{\left(0,7\right)^2}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2 : Bài giải
\(a,\text{ }\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3\)
\(=\frac{9}{10}-\frac{7}{10}+9,3=\frac{1}{5}+9,3=0,2+9,3=9,5\)
\(b,\text{ }\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)
\(c,\text{ }\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)
Bài 2 :
a ) \(\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3\)
\(=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3=\frac{9}{10}-\frac{7}{10}+9,3\)
\(=\frac{1}{5}+9,3=0,2+9,3=9,5\)
b ) \(\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}\)
\(=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)
c ) \(\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}\)
\(=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bai 1
a) \(\sqrt{0,36}+\sqrt{0,49}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\frac{2}{3}-\frac{5}{6}\)
=\(-\frac{1}{6}\)
Bài 2
a)\(x^2=81\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
b) \(\left(x-1\right)^2=\frac{9}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{3}{4}\\x-1=\frac{-3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
c) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{\frac{36}{81}}+\sqrt{\frac{16}{25}}\right):\sqrt{\frac{100}{36}}=\left[\sqrt{\left(\frac{6}{9}\right)^2}+\sqrt{\left(\frac{4}{5}\right)^2}\right]:\sqrt{\left(\frac{10}{6}\right)^2}\)
=\(\left(\frac{6}{9}+\frac{4}{5}\right):\frac{10}{6}=\left(\frac{2}{3}+\frac{4}{5}\right).\frac{3}{5}=\left(\frac{10}{15}+\frac{12}{15}\right).\frac{3}{5}=\frac{22}{15}.\frac{3}{5}=\frac{22}{25}\)
Ta có:
\(\sqrt{0,49}+\sqrt{\frac{25}{36}}=\sqrt{0,7^2}+\sqrt{\left(\frac{5}{6}\right)^2}=0,7+\frac{5}{6}=\frac{23}{15}\)