Đáp án: $x=\dfrac12$
Giải thích các bước giải:
c.ĐKXĐ: $x>-\dfrac12, x\ne 0$
Ta có:
$\dfrac{x}{\sqrt{2x+1}}+\dfrac{\sqrt{2x+1}}{8x}=\dfrac{\sqrt{2}}{2}$
Vì $\dfrac{\sqrt{2}}{2}>0\to x>0$
$\to \dfrac{x}{\sqrt{2x+1}},\dfrac{\sqrt{2x+1}}{8x}>0$
$\to \dfrac{x}{\sqrt{2x+1}}+\dfrac{\sqrt{2x+1}}{8x}\ge 2\sqrt{\dfrac{x}{\sqrt{2x+1}}\cdot\dfrac{\sqrt{2x+1}}{8x}}$
$\to \dfrac{x}{\sqrt{2x+1}}+\dfrac{\sqrt{2x+1}}{8x}\ge \dfrac{\sqrt{2}}{2}$
$\to$Dấu = xảy ra khi:
$\dfrac{x}{\sqrt{2x+1}}=\dfrac{\sqrt{2x+1}}{8x}$
$\to 8x^2=2x+1$
$\to 8x^2-2x-1=0$
$\to (2x-1)(4x+1)=0$
Mà $x>0\to x=\dfrac12$
