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pt1$\Leftrightarrow4x^3-6x^2y+3xy^2-y^3+x-y=0$
$\Leftrightarrow (x-y)^3+3x^2(x-y)+(x-y)=0$
$\Leftrightarrow (x-y)[(x-y)^2+3x^2+1]=0$
$\Leftrightarrow \left[\begin{gathered}x=y\\(x-y)^2+3x^2+1=0(\ast)\\ \end{gathered} \right.$
dễ thấy pt(*) vô nghiệm. vậy x=y
thay vào pt 2 có: $\sqrt{3x+1}+\sqrt[3]{x+7}=4$
$\Leftrightarrow (\sqrt{3x+1}-2)+(\sqrt[3]{x+7}-2)=0$
$\Leftrightarrow \frac{3x-3}{2+\sqrt{3x+1}}+\frac{x-1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}=0$
$\Leftrightarrow \left[\begin{gathered}x=1\\\frac{3}{2+\sqrt{3x+1}}+\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}=0(3)\\ \end{gathered} \right.$
pt 3: VT>0$\Rightarrow $ pt vô nghiệm
$\Rightarrow $ x=y=1
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