$\sqrt{2x^3+6}=x+\sqrt{x^2-3x+3}$
Giải phương trình $\sqrt{2x^3+6}=x+\sqrt{x^2-3x+3}$
Started By trang2004, 12-09-2018 - 12:04
#1
Posted 12-09-2018 - 12:04
#2
Posted 19-09-2018 - 22:38
$\sqrt{2x^3+6}=x+\sqrt{x^2-3x+3}$
Xin được sửa đề thành $\sqrt[3]{2x^3+6}=x+\sqrt{x^2-3x+3}$
Pt đã cho
$<=>\sqrt[3]{2x^3+6}-(\frac{2-\sqrt[3]{\frac{9}{2}}}{1+\sqrt[3]{\frac{3}{4}}}x+\frac{\sqrt[3]{\frac{9}{2}}+2\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}})=\sqrt{x^2-3x+3}-(\frac{1-\sqrt[3]{\frac{9}{2}}-\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}}x+\frac{\sqrt[3]{\frac{9}{2}}+2\sqrt[3]{\frac{3}{4}}}{1+\sqrt[3]{\frac{3}{4}}})$
$<=>...<=>(x+\sqrt[3]{\frac{3}{4}})(x-1).A=0($Với $A\neq 0\forall x\in \mathbb{R})$
$<=>x=-\sqrt[3]{\frac{3}{4}};x=1$
Vậy$,...$
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