Chủ đề này có 1 trả lời

[19kvh97]

Giải pt
$\frac{3+\sqrt{x}}{x^2+x\sqrt{x}+x+3}+\frac{x+\sqrt{x}+2}{x^2+x\sqrt{x}+4}+\frac{x\sqrt{x}+x+2}{x^2+\sqrt{x}+4}+\frac{x^2+x\sqrt{x}+2}{x+\sqrt{x}+4}+\frac{x^2+3}{x\sqrt{x}+x+\sqrt{x}+3}=\frac{10}{3}$

[phanquockhanh]

Giải pt
$\frac{3+\sqrt{x}}{x^2+x\sqrt{x}+x+3}+\frac{x+\sqrt{x}+2}{x^2+x\sqrt{x}+4}+\frac{x\sqrt{x}+x+2}{x^2+\sqrt{x}+4}+\frac{x^2+x\sqrt{x}+2}{x+\sqrt{x}+4}+\frac{x^2+3}{x\sqrt{x}+x+\sqrt{x}+3}=\frac{10}{3}$(1)

Đặt a=2,$b=\sqrt{x}+1,c=x+1,d=x\sqrt{x}+1,e=1+x^{2}$
$(1)\Leftrightarrow \frac{a+b}{c+d+e}+\frac{b+c}{d+e+a}+\frac{c+d}{e+a+b}+\frac{d+e}{a+b+c}+\frac{e+a}{b+c+d}= \frac{10}{3}$
$\Leftrightarrow \frac{a+b}{c+d+e}+1+\frac{b+c}{d+e+a}+1+\frac{c+d}{e+a+b}+1+\frac{d+e}{a+b+c}+1+\frac{e+a}{b+c+d}+1= \frac{10}{3}+5$
$\Leftrightarrow\left ( a+b+c+d +e\right )\left ( \frac{1}{c+d+e}+\frac{1}{d+e+a} +\frac{1}{e+a+b}+\frac{1}{a+b+c}+\frac{1}{b+c+d}\right )=\frac{25}{3}$
$\Leftrightarrow ((c+d+e)+(d+e+a)+(e+a+b)+(a+b+c)+(b+c+d) ) ( \frac{1}{c+d+e}+\frac{1}{d+e+a} +\frac{1}{e+a+b}+\frac{1}{a+b+c}+\frac{1}{b+c+d} )={25}$
Theo AM-GM ta có:$VT\geq 25$
Dấu "=" xảy ra $\Leftrightarrow a=b=c=d=e\Leftrightarrow x=1$