1,$\sqrt[3]{3x+2}+x\sqrt{3x-2}=2\sqrt{2x^{2}+1}$
2,$x^{2}-5x+2\sqrt{3x+12}=0$
$\sqrt[3]{3x+2}-2 +x(\sqrt{3x-2}-2) = 2(\sqrt{2x^2 +1}-(x+1))$ (DK: $x\geq \frac{2}{3}$)
$\Leftrightarrow \frac{3(x-2)}{(\sqrt[3]{3x+2})^2+2\sqrt[3]{3x+2}+4} + \frac{3x(x-2)}{\sqrt{3x-2}+2} =\frac{2x(x-2)}{\sqrt{2x^2+1}+x+1}$
$\Leftrightarrow x=2$ Thỏa mãn.
$\Leftrightarrow \frac{3}{(\sqrt[3]{3x+2})^2+2\sqrt[3]{3x+2}+4} +\frac{3x}{\sqrt{3x-2}+2} -\frac{2x}{\sqrt{2x^2+1}+x+1}=0 (*)$
Xet : $\frac{3}{\sqrt{3x-2}+2} -\frac{2}{\sqrt{2x^2+1}+x+1} \Leftrightarrow \frac{3\sqrt{2x^2+1}+(3x-2)-2\sqrt{3x-2}+1}{...} = \frac{3\sqrt{2x^2+1}+(\sqrt{3x-2}-1)^2}{...} >0$ ( $\forall x\geq \frac{2}{3}$)
$\Rightarrow pt(*)$ có $VT>0=VP \Rightarrow (*)$ Vô nghiệm.
Bài viết đã được chỉnh sửa nội dung bởi didifulls: 14-09-2017 - 19:23