daovantest 31. Giải hệ phương trình: \left\{\begin{array}c\frac{2xy}{x+y}\sqrt{\frac{x^2+y^2}2}=\frac{2\sqrt{xy}+x+y}2\\sqrt[3]{9xy+3x+6y+9}+2\sqrt[3]{6xy+2}\end{array} ight. \sqrt[3]{9xy+3x+6y+9}+2\sqrt[3]{6xy+2}=3x+4

Question

daovantest 31.  Giải hệ phương trình:   \left\{\begin{array}c\frac{2xy}{x+y}\sqrt{\frac{x^2+y^2}2}=\frac{2\sqrt{xy}+x+y}2\\sqrt[3]{9xy+3x+6y+9}+2\sqrt[3]{6xy+2}\end{array}ight.   \sqrt[3]{9xy+3x+6y+9}+2\sqrt[3]{6xy+2}=3x+4

Accepted Answer

Sửa lại đề $\displaystyle \begin{cases}
\dfrac{2xy}{x+y} +\sqrt{\dfrac{x^{2} +y^{2}}{2}} =\dfrac{2\sqrt{xy} +x+y}{2} \ ( 1)\
\sqrt[3]{9xy+3x+6y+9} +2\sqrt[3]{6xy+2} =3x+4\ ( 2)
\end{cases}$
Đk:$\displaystyle xy\geqslant 0;x+y eq 0$
Ta có $\displaystyle ( 1) \Leftrightarrow \left(\dfrac{2xy}{x+y} -\dfrac{x+y}{2} ight) +\left(\sqrt{\dfrac{x^{2} +y^{2}}{2}} -\sqrt{xy} ight) =0$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow -\dfrac{( x-y)^{2}}{x+y} +\dfrac{( x-y)^{2}}{\left(\sqrt{\dfrac{x^{2} +y^{2}}{2}} +\sqrt{xy} ight)} =0\
\Leftrightarrow \left[ \begin{array}{l l}
x=y\
x+y=\sqrt{\dfrac{x^{2} +y^{2}}{2}} +\sqrt{xy}
\end{array} ight.
\end{array}$
+/Nếu $\displaystyle x=y$, thay vào $\displaystyle ( 2)$ ta được $\displaystyle \sqrt[3]{9x^{2} +9x+9} +2\sqrt[3]{6x^{2} +2} =3x+4$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \left(\sqrt[3]{9x^{2} +9x+9} -( x+2) ight) +2\left(\sqrt[3]{6x^{2} +2} -( x+1) ight) =0\
\Leftrightarrow \dfrac{-( x-1)^{3}}{A^{2} +( x+2) A+( x+2)^{2}} +2\dfrac{-( x-1)^{3}}{B^{2} +( x+1) B+( x+1)^{2}} =0
\end{array}$
với $\displaystyle A=\sqrt[3]{9x^{2} +9x+9} ;\ B=\sqrt[3]{6x^{2} +2}$
$\displaystyle \Leftrightarrow x=1\Rightarrow y=1\ ( tm)$ (do $\displaystyle A^{2} +( x+2) A+( x+2)^{2} >0;B^{2} +( x+1) B+( x+1)^{2} >0)$
+/Nếu $\displaystyle x+y=\sqrt{\dfrac{x^{2} +y^{2}}{2}} +\sqrt{xy} \ ( x;y\geqslant 0)$
Áp dụng bđt CBS ta có $\displaystyle \left(\sqrt{\dfrac{x^{2} +y^{2}}{2}} +\sqrt{xy} ight)^{2} \leqslant 2\left(\dfrac{x^{2} +y^{2}}{2} +xy ight) =( x+y)^{2}$
$\displaystyle \Rightarrow x+y=\sqrt{\dfrac{x^{2} +y^{2}}{2}} +\sqrt{xy} \leqslant \sqrt{( x+y)^{2}} =x+y$
Dấu bằng xảy ra $\displaystyle \Leftrightarrow x=y$
Thay vào $\displaystyle ( 2)$ ta cũng được $\displaystyle x=y=1$