mn giúp mk vs Cho x=$\sqrt[3]{3+2\sqrt{2}}$ +$\sqrt[3]{3-2\sqrt{2}}$ y=$\sqrt[3]{17+12\sqrt{2}}$ +$\sqrt[3]{17-12\sqrt{2}}$ Tính A=$(x-y)^{2}$

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1. Đáp án: (Mình đổi đề: $(x-y)^2⇒(x-y)^3;xy+1⇒xy-1$ nhé)

   $A=-28$ 

   Giải thích các bước giải:

   Ta có:

   $x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}$

   $⇒x^3=(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}})^3$

   $=3+2\sqrt{2}+3-2\sqrt{2}+3.\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}.(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}})$

   $=6+3x.\sqrt[3]{(3+2\sqrt{2})(3-2\sqrt{2})}$

   $=6+3x.\sqrt[3]{9-8}$

   $=6+3x.\sqrt[3]{1}=6+3x$

   $⇒x^3-3x=6$

   $y=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}$

   $⇒y^3=(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}})^3$

   $=17+12\sqrt{2}+17-12\sqrt{2}+3.\sqrt[3]{17+12\sqrt{2}}.\sqrt[3]{17-12\sqrt{2}}.(\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}})$

   $=34+3y.\sqrt[3]{(17+12\sqrt{2})(17-12\sqrt{2})}$

   $=34+3y.\sqrt[3]{289-288}$

   $=34+3y.\sqrt[3]{1}=34+3y$

   $⇒y^3-3y=34$

   Lại có:

   $A=(x-y)^3+3(x-y)(xy-1)$

   $=(x^3-3x^2y+3xy^2-y^3)+(3x^2y-3x-3xy^2+3y)$

   $=x^3-3x^2y+3xy^2-y^3+3x^2y-3x-3xy^2+3y$

   $=x^3-3x-y^3+3y$

   $=(x^3-3x)-(y^3-3y)$

   $=6-34=-28$

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