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[studentlovemath]

Giải hệ phương trình: $\left\{\begin{matrix}y^{2}+(4x-1)^{2}=\sqrt[3]{4x(8x+1)}\\ 40x^{2}+x=y\sqrt{14x-1}\end{matrix}\right.$

[Baoriven]

Ta có: $80x^{2}+2x=2y\sqrt{14x-1}\leq y^{2}+14x-1\Rightarrow y^{2}\geq 80x^{2}-12x+1$.

$\Rightarrow \sqrt[3]{4x(8x+1)}\geq 96x^{2}-20x+2$.

Ta lại có $\sqrt[3]{4x(8x+1)}=2\sqrt[3]{\frac{1}{2}.4x.\frac{8x+1}{4}}\leq \frac{2}{3}(\frac{1}{2}+4x+\frac{8x+1}{4})\leq 96x^2-20x+2$.

Nên $x=\frac{1}{8}$.

Từ đó suy ra $y=\frac{\sqrt{3}}{2}$.

Edited by Baoriven, 03-03-2017 - 19:01.