B1:thực hiện phép tính:
a. [căn(9/20 - căn(1/2)] x căn 2
b.(căn12 + căn27 - căn3) x căn3
c.( căn(8/3) - căn24 + căn(50/3) ) x căn6
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1, \(sin\left(x+\dfrac{\pi}{6}\right)+cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{6}}{2}\)
⇔ \(\dfrac{\sqrt{2}}{2}sin\left(x+\dfrac{\pi}{6}\right)+\dfrac{\sqrt{2}}{2}cos\left(x+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)
⇔ \(sin\left(x+\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)=sin\dfrac{\pi}{4}\)
2, \(\left(\sqrt{3}-1\right)sinx+\left(\sqrt{3}+1\right)cosx=1-\sqrt{3}\)
⇔ \(\dfrac{\left(\sqrt{3}-1\right)}{2\sqrt{2}}sinx+\dfrac{\left(\sqrt{3}+1\right)}{2\sqrt{2}}cosx=\dfrac{1-\sqrt{3}}{2\sqrt{2}}\)
⇔ sinx . si
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\dfrac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}\)
\(=\dfrac{\sqrt{10}\left(2+\sqrt{3}\right)-\sqrt{2}\left(2+\sqrt{3}\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{5}-1\right)}{2\sqrt{2}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2+\sqrt{3}}{2}\)
b) Ta có: \(\sqrt{\left(1-\sqrt{2006}\right)^2}\cdot\sqrt{2007+2\sqrt{2006}}\)
\(=\left(\sqrt{2006}-1\right)\left(\sqrt{2006}+1\right)\)
=2005
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)
\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)
\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)
\(=\dfrac{3\sqrt{10}}{10}-1\)
b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)
\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)
c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)
\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)