1, Có $x^{2}+2ax+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}}$1) $x^2 + 2ax +\frac{1}{16}=-a+\sqrt{a^2+x -\frac{1}{16}}$ với $a\epsilon \left ( 0;\frac{1}{4} \right )$
$\Leftrightarrow (a^{2}+2ax+x^{2})-(a^{2}+x-\frac{1}{16})+(a+x)=\sqrt{a^{2}+x-\frac{1}{16}}$
$\Leftrightarrow (x+a)^{2}-(a^{2}+x-\frac{1}{16})+(x+a)-\sqrt{a^{2}+x-\frac{1}{16}}=0$
$\Leftrightarrow ((x+a)-\sqrt{a^{2}+x-\frac{1}{16}})((x+a)+\sqrt{a^{2}+x-\frac{1}{16}}+1)=0$
Th1:$x+a=\sqrt{a^{2}+x-\frac{1}{16}}$
Đk: $x\geq -a$
$\Rightarrow x^{2}+x(2a-1)+\frac{1}{16}=0$
$\Leftrightarrow x=...$
Th2:$\sqrt{a^{2}+x-\frac{1}{16}}=-x-a-1$
Đk: $x\leq -a-1$
$\Rightarrow x^{2}+x(2a+1)+2a+\frac{17}{16}=0$
$\Rightarrow x=...$