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Đã tìm ra cách giải! 
$\int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{2x-\sqrt{4x^{2}-1}}dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{2x-\sqrt{4(x-\frac{1}{2})(x+\frac{1}{2})}}dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{2x-2\sqrt{(x-\frac{1}{2})(x+\frac{1}{2})}}dx$
Nhận thấy bên trong căn có dạng:
$(A\sqrt{x+\frac{1}{2}}+B\sqrt{x-\frac{1}{2}})^{2}$
$\Leftrightarrow (A^{2}+B^{2})x + \frac{1}{2}(A^{2}-B^{2}) + 2AB\sqrt{(x+\frac{1}{2})(x-\frac{1}{2})}$
$\Leftrightarrow \left\{\begin{matrix}A^{2} + B^{2} = 2 & & \\ A^{2} - B^{2} = 0 & & \\ AB = -1 & & \end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix} A = 1 & \\ B = -1 & \end{matrix}\right.v \left\{\begin{matrix} A = -1 & \\ B = 1 & \end{matrix}\right.$
Với $A = 1, B = -1$
$\int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{2x-\sqrt{4x^{2}-1}}dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{(\sqrt{x+\frac{1}{2}}-\sqrt{x-\frac{1}{2}})^{2}}dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} {\sqrt{x+\frac{1}{2}}-\sqrt{x-\frac{1}{2}}}dx$
= $\frac{2}{3}(x-\frac{1}{2})\sqrt{x-\frac{1}{2}}+\frac{2}{3}(x+\frac{1}{2})\sqrt{x+\frac{1}{2}}\left.\begin{matrix}\frac{3}{2} & \\ \frac{1}{2} & \end{matrix}\right|$
$= \frac{4}{3} (\sqrt{2}-1)$
Với $A = -1, B = 1$
$\int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{2x-\sqrt{4x^{2}-1}}dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} \sqrt{(\sqrt{x-\frac{1}{2}}- \sqrt{x+\frac{1}{2}})^{2} }dx$
$= \int_{\frac{1}{2}}^{\frac{3}{2}} {\sqrt{x-\frac{1}{2}} - \sqrt{x+\frac{1}{2}}}dx$
= $\frac{2}{3}(x-\frac{1}{2})\sqrt{x-\frac{1}{2}}-\frac{2}{3}(x+\frac{1}{2})\sqrt{x+\frac{1}{2}}\left.\begin{matrix}\frac{3}{2} & \\ \frac{1}{2} & \end{matrix}\right|$
$= -\frac{4}{3} (\sqrt{2}-1)$ (loại)
Vậy, đáp số là $= \frac{4}{3} (\sqrt{2}-1)$
- lmht và uahnbu29main thích
