Đk $x \le \frac 52 $$pt\Leftrightarrow 4x^3+x=(3-x)\sqrt{5-2x}$$\Leftrightarrow4x^3+x=\frac 12[(5-2x)\sqrt{5-2x}+\sqrt{5-2x}]$$\Leftrightarrow 4x^3+x=\frac {\sqrt{5-2x}^3}2+\frac{\sqrt{5-2x}}2$$\Leftrightarrow 4x^3+x=4.\left(\frac{\sqrt{5-2x}}{2}\right)^3+\left( \frac{5-2x}2 \right)$Xét $f(t)=4t^3+t$ có $f'(t)=12t^2+1>0\Rightarrow f(t)$ đồng biến trên $\mathbb{R}$Nên $f(x)=f(\frac{\sqrt{5-2x}}{2})$$\Leftrightarrow x=\frac{\sqrt{5-2x}}{2}$$\Leftrightarrow \begin{cases} \frac 52\ge x \ge 0 \\ 4x^2=5-2x \end{cases}\Leftrightarrow \boxed{x=\frac{-1+\sqrt {21}}{4}}$
Đk $x \le \frac 52 $$pt\Leftrightarrow 4x^3+x=(3-x)\sqrt{5-2x}$$\Leftrightarrow4x^3+x=\frac 12[(5-2x)\sqrt{5-2x}+\sqrt{5-2x}]$$\Leftrightarrow 4x^3+x=\frac {\sqrt{5-2x}^3}2+\frac{\sqrt{5-2x}}2$$\Leftrightarrow 4x^3+x=4.\left(\frac{\sqrt{5-2x}}{2}\right)^3+\left( \frac{5-2x}2 \right)$Xét $f(t)=4x^3+x$ có $f'(t)=12x^2+1>0\Rightarrow f(t)$ đồng biến trên $\mathbb{R}$Nên $f(x)=f(\frac{\sqrt{5-2x}}{2})$$\Leftrightarrow x=\frac{\sqrt{5-2x}}{2}$$\Leftrightarrow \begin{cases} \frac 52\ge x \ge 0 \\ 4x^2=5-2x \end{cases}\Leftrightarrow \boxed{x=\frac{-1+\sqrt {21}}{4}}$
Đk $x \le \frac 52 $$pt\Leftrightarrow 4x^3+x=(3-x)\sqrt{5-2x}$$\Leftrightarrow4x^3+x=\frac 12[(5-2x)\sqrt{5-2x}+\sqrt{5-2x}]$$\Leftrightarrow 4x^3+x=\frac {\sqrt{5-2x}^3}2+\frac{\sqrt{5-2x}}2$$\Leftrightarrow 4x^3+x=4.\left(\frac{\sqrt{5-2x}}{2}\right)^3+\left( \frac{5-2x}2 \right)$Xét $f(t)=4
t^3+
t$ có $f'(t)=12
t^2+1>0\Rightarrow f(t)$ đồng biến trên $\mathbb{R}$Nên $f(x)=f(\frac{\sqrt{5-2x}}{2})$$\Leftrightarrow x=\frac{\sqrt{5-2x}}{2}$$\Leftrightarrow \begin{cases} \frac 52\ge x \ge 0 \\ 4x^2=5-2x \end{cases}\Leftrightarrow \boxed{x=\frac{-1+\sqrt {21}}{4}}$