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Câu hỏi: Cho $m$, $n$ là hai số dương không đồng thời bằng $1$, biểu thức $\dfrac{{{m}^{2\sqrt{2}}}-{{n}^{2\sqrt{3}}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}-1$ bằng
A. $\dfrac{2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
B. $\dfrac{-2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
C. $\dfrac{2{{m}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
D. $\dfrac{-2{{m}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
A. $\dfrac{2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
B. $\dfrac{-2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
C. $\dfrac{2{{m}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
D. $\dfrac{-2{{m}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
Ta có: $\dfrac{{{m}^{2\sqrt{2}}}-{{n}^{2\sqrt{3}}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}-1=\dfrac{{{m}^{2\sqrt{2}}}-{{n}^{2\sqrt{3}}}-{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}=\dfrac{{{m}^{2\sqrt{2}}}-{{n}^{2\sqrt{3}}}-{{m}^{2\sqrt{2}}}-{{n}^{2\sqrt{3}}}+2{{m}^{\sqrt{2}}}{{n}^{\sqrt{3}}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}$
$=\dfrac{-2{{n}^{\sqrt{3}}}+2{{m}^{\sqrt{2}}}{{n}^{\sqrt{3}}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}=\dfrac{2{{n}^{\sqrt{3}}}\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}=\dfrac{2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
$=\dfrac{-2{{n}^{\sqrt{3}}}+2{{m}^{\sqrt{2}}}{{n}^{\sqrt{3}}}}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}=\dfrac{2{{n}^{\sqrt{3}}}\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}{{{\left( {{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}} \right)}^{2}}}=\dfrac{2{{n}^{\sqrt{3}}}}{{{m}^{\sqrt{2}}}-{{n}^{\sqrt{3}}}}$.
Đáp án A.