Question

[Phương Anh]

\(\begin{cases}\left(18x+9\right)\sqrt{x^2+x+1}=y\sqrt{4y^2+27}\\\left(2y+3\right)^2=24\sqrt{x}\left(2y-9\right)\end{cases}\)

Answer 1

[Huỳnh Tâm]

ĐK: \(x\ge0;y\ge\frac{9}{2}\)

(1) \(\Leftrightarrow6\left(x+\frac{1}{2}\right)\sqrt{\left[3\left(x+\frac{1}{2}\right)\right]^2+\frac{27}{4}}=2y\sqrt{y^2+\frac{27}{4}}\)

Xét \(f\left(t\right)=2t\sqrt{t^2+\frac{27}{4}}\left(t>0\right)\)

\(f'\left(t\right)=2\sqrt{t^2+\frac{27}{4}}+\frac{2t^2}{\sqrt{t^2+\frac{27}{4}}}>0;\forall t>0\)

→ hàm đồng biến trên (0;+∞)

Mà \(f\left(3\left(x+\frac{1}{2}\right)\right)=f\left(y\right)\Leftrightarrow3\left(x+\frac{1}{2}\right)=y\)

Thế vào (2) ta được: 

\(\left(6y+6\right)^2=24\sqrt{x}\left(6y-6\right)\Leftrightarrow\left(x+1\right)^2=4\sqrt{x}\left(x-1\right)\)

\(\Leftrightarrow\left(\sqrt{x}\right)^4-4\left(\sqrt{x}\right)^3+2\left(\sqrt{x}\right)^2+4\sqrt{x}+1=0\)

\(\Leftrightarrow\left(\sqrt{x}\right)^4+4\sqrt{x}+1-2\cdot x\cdot2\sqrt{x}-2\cdot x\cdot1+2\cdot1\cdot2\sqrt{x}=0\)

\(\Leftrightarrow\left(x-2\sqrt{x}-1\right)^2=0\)

\(\Leftrightarrow x-2\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1+\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)

\(\Rightarrow y=\frac{21+12\sqrt{2}}{2}\)