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tính giá trị của biểu thức \(\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right)\sqrt{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}.\)
\(=\left(\frac{\sqrt{3}}{\sqrt{4}}-\sqrt{3}+5\cdot\frac{\sqrt{4}}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(\frac{\sqrt{3}}{2}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(\frac{1}{2}\cdot\sqrt{3}-\sqrt{3}+5\cdot\frac{2}{\sqrt{3}}\right)\sqrt{12}.\)
\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{\sqrt{3}}\right)\sqrt{12}\)
\(=\left(-\frac{1}{2}\sqrt{3}+\frac{10}{3}\sqrt{3}\right)\sqrt{12}\)
\(=\frac{17}{6}\sqrt{3}\sqrt{12}=\frac{17}{6}\sqrt{36}=\frac{17}{6}\cdot6=17\)
\(\Rightarrow\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\frac{4}{3}}\right)\sqrt{12}=17\)
![](https://rs.olm.vn/images/avt/0.png?1311)
NX \(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}\) =\(\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}-1\right)}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}+1\right)^2}\)
=\(\frac{\left(\left(\sqrt{n+1}-\sqrt{n}\right)^2-1^2\right)}{n+1-n-1-2\sqrt{n}}\) \(=\frac{n+1+n-2\sqrt{\left(n+1\right)n}-1}{-2\sqrt{n}}=\frac{2n-2\sqrt{n\left(n+1\right)}}{-2\sqrt{n}}\)
=\(\frac{n-\sqrt{n\left(n+1\right)}}{-\sqrt{n}}=\frac{n}{-\sqrt{n}}+\frac{\sqrt{n\left(n+1\right)}}{\sqrt{n}}=-\sqrt{n}+\sqrt{n+1}\)
thay vao Q ta co
Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)