Giải phương trình:
$\frac{3+\sqrt{x}}{x^{2}+x\sqrt{x}+x+3}+\frac{x+\sqrt{x}+2}{x^{2}+x\sqrt{x}+3}+\frac{x\sqrt{x}+x+2}{x^{2}+\sqrt{x}+4}+\frac{x^{2}+x\sqrt{x}+2}{x+x\sqrt{x}+4}+\frac{x^{2}+3}{x\sqrt{x}+x+\sqrt{x}+3}=\frac{10}{3}$
ĐK: $x\geq 0$
Đặt $a=2, b=\sqrt{x}+1, c=x+1, d=x\sqrt{x}+1, e=x^{2}+1$
Khi đó phương trình đã cho trở thành:
$\frac{a+b}{c+d+e}+\frac{b+c}{d+e+a}+\frac{c+d}{e+a+b}+\frac{d+e}{a+b+c}+\frac{e+a}{b+c+d}=\frac{10}{3}$
$\Leftrightarrow \left ( \frac{a+b}{c+d+e}+1 \right )+\left ( \frac{b+c}{d+e+a}+1 \right )+\left ( \frac{c+d}{e+a+b}+1 \right )+\left ( \frac{d+e}{a+b+c}+1 \right )+\left ( \frac{e+a}{b+c+d}+1 \right )=\frac{25}{3}$
$\Leftrightarrow (a+b+c+d+e)\left ( \frac{1}{c+d+e}+\frac{1}{d+e+a}+\frac{1}{e+a+b}+\frac{1}{a+b+c}+\frac{1}{b+c+d} \right )=\frac{25}{3}$
$\Leftrightarrow \left [ (c+d+e)+(d+e+a)+(e+a+b)+(a+b+c)+(b+c+d) \right ]\left ( \frac{1}{c+d+e}+\frac{1}{d+e+a}+\frac{1}{e+a+b}+\frac{1}{a+b+c}+\frac{1}{b+c+d} \right )=25$(*)
Áp dụng bất đẳng thức AM-GM ta có: $VT_{*}\geq 25$
Dấu = xảy ra$\Leftrightarrow a=b=c=d=e\Leftrightarrow x=1$