Chứng minh rằng $\frac{2\sqrt{mn} }{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}$=$\sqrt{m}$+$\sqrt{n}$-$\sqrt{m+n}$ 01/07/2021 Bởi Allison Chứng minh rằng $\frac{2\sqrt{mn} }{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}$=$\sqrt{m}$+$\sqrt{n}$-$\sqrt{m+n}$
$\dfrac{2\sqrt{mn} }{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}$ $=\dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})(\sqrt{m}+\sqrt{n}-\sqrt{m+n})}$ $= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{(\sqrt m + \sqrt n)^2 – (\sqrt{m+n})^2}$ $= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{m + n + 2\sqrt{mn}- (m + n)}$ $= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{2\sqrt{mn}}$ $= \sqrt{m}+\sqrt{n}-\sqrt{m+n} \quad (đpcm)$ Bình luận
$\dfrac{2\sqrt{mn} }{\sqrt{m}+\sqrt{n}+\sqrt{m+n}}$
$=\dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{(\sqrt{m}+\sqrt{n}+\sqrt{m+n})(\sqrt{m}+\sqrt{n}-\sqrt{m+n})}$
$= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{(\sqrt m + \sqrt n)^2 – (\sqrt{m+n})^2}$
$= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{m + n + 2\sqrt{mn}- (m + n)}$
$= \dfrac{2\sqrt{mn}(\sqrt{m}+\sqrt{n}-\sqrt{m+n}) }{2\sqrt{mn}}$
$= \sqrt{m}+\sqrt{n}-\sqrt{m+n} \quad (đpcm)$