HPT!

Question

$\begin{cases}y^{2}+(4x-1)^{2}=\sqrt[3]{4x(8x+1)} \\ 40x^{2}+x=y\sqrt{14x-1}\end{cases}$

★★.P.I.N.O.★★ · Answer 1 · 3/21/2016 9:05:56 AM

$\color{green}{\begin{cases}y^{2}+(4x-1)^{2}=\sqrt[3]{4x(8x+1)} \\ 40x^{2}+x=y\sqrt{14x-1}\end{cases}}$

Điều kiện:

$x \ge \frac{1}{14}.$

Lấy PT

$(1)+2.$ PT

$(2),$ ta được:

$y^{2}+(4x-1)^{2}+2(40x^{2}+x)=\sqrt[3]{4x(8x+1)}+2y\sqrt{14x-1}$

$(1)$

Mặt khác, ta có:

$\sqrt[3]{4x(8x+1)}+2y\sqrt{14x-1}=\sqrt[3]{8x.\frac{8x+1}{2}.1}+2y\sqrt{14x-1}$

$\le \frac{1}{3}(8x+\frac{8x+1}{2}+1)+y^2+14x-1=y^2+(4x-1)^2+2(40x^2+x)-\frac{3}{2}(8x-1)^2$

$\le y^2+(4x-1)^2+2(40x^2+x)$

$(2)$

Từ

$(1),(2)$ suy ra đẳng thức xảy ra

$\Leftrightarrow x=\frac{1}{8};y=\frac{\sqrt3}{2}.$

Kết luận: hệ đã cho có nghiệm duy nhất

$\color{red}{(x;y)=(\frac{1}{8};\frac{\sqrt3}{2})}$

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