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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a\)
\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)
\(=\)\(\sqrt{2,7.1,2}\)
\(=\)\(\sqrt{3,24}\)
\(=\)\(1,8\)
\(b\)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)
\(=\)\(\sqrt{85.125.68}\)
\(=\)\(\sqrt{722500}\)
\(=\)\(850\)
học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\sqrt{13,5}}{\sqrt{4,5}}=\sqrt{\frac{13,5}{4,5}}=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}=\sqrt{85.125.68}=\sqrt{5.17.5.25.17.4}\)
\(=\sqrt{5^2.25.17^2.4}=\sqrt{5^2}.\sqrt{25}.\sqrt{17^2}.\sqrt{4}=5.5.17.2=850\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)
\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)
Vậy A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{2,5.2560}=\sqrt{25.256}=\sqrt{25}.\sqrt{256}=5.16=80\)
b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{\frac{7}{2}}.\sqrt{\frac{5}{2}}.\sqrt{7}.\sqrt{\frac{1}{5}}\)
\(=\sqrt{\frac{7}{2}.\frac{5}{2}.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)
c) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1}.\sqrt{0,09}\)
\(=\sqrt{4.121}.\sqrt{9.0,01}=\sqrt{4}.\sqrt{121}.\sqrt{9}.\sqrt{0,01}\)
\(=2.11.3.0,1=6,6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn hình như chép nhầm đề rồi
\(4=\sqrt{16};2\sqrt{15}=\sqrt{60}\)
Sao trừ trong căn được
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\left(\sqrt{3+\sqrt{20}}\right)^2-\left(\sqrt{5+\sqrt{5}}\right)^2\)
\(=3+\sqrt{20}-5-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
Ta thấy: \(5>4\Rightarrow\sqrt{5}>\sqrt{4}\Rightarrow\sqrt{5}>2\)
Do đó : hiệu trên >0
Suy ra : \(\sqrt{3+\sqrt{20}}>\sqrt{5+\sqrt{5}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
ĐKXĐ \(\hept{\begin{cases}x\ge0\\x-1\ne0\\x+2\sqrt{x}+1\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne1\\\left(\sqrt{x}+1\right)^2\ne0\end{cases}}\)
\(Q=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)
\(=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\frac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}\)
\(=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\sqrt{x}=\frac{2x}{x-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nhập PT vào máy tính, sử dụng dầu "=" ô nút CALC.
sau khi nhập xong, nhấn SHIFT,CALC, rồi nhấn dấu =
Ta được x=-1,322875656
Ta có: \(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{81}{25}}=\sqrt{\left(\frac{9}{5}\right)^2}=\frac{9}{5}\)
\(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{27}{5}\cdot\frac{3}{5}}\)
\(=\sqrt{\frac{81}{25}}\)
\(=\sqrt{\left(\frac{9}{5}\right)^2}\)
\(=\left|\frac{9}{5}\right|=\frac{9}{5}\)