Đáp án: $m\ge \dfrac34$ hoặc $m\le-\dfrac34$
Giải thích các bước giải:
Ta có:
$(x^2+\dfrac{1}{x^2})-2m(x+\dfrac1x)+1=0$
$\to (x^2+2x\cdot\dfrac1x+\dfrac{1}{x^2})-2m(x+\dfrac1x)-1=0$
$\to (x+\dfrac1x)^2-2m(x+\dfrac1x)-1=0$
$\to 2m(x+\dfrac1x)=(x+\dfrac1x)^2-1$
Nếu $x+\dfrac1x=0$
$\to 2\cdot 0=0^2-1$ vô lý
$\to x+\dfrac1x\ne 0$
$\to 2m=\dfrac{(x+\dfrac1x)^2-1}{x+\dfrac1x}$
Đặt $x+\dfrac1x=t, t\ne 0, |t|\ge 2$
$\to 2m=\dfrac{t^2-1}{t}$
Nếu $t\ge 2$
Xét hàm số $y=\dfrac{t^2-1}{t}$
$\to y=t-\dfrac1t\ge 2-\dfrac12=\dfrac32$
$\to 2m\ge \dfrac32$
$\to m\ge \dfrac34$
Nếu $t\le -2$
$\to y=t-\dfrac1t\le -2-\dfrac1{-2}=-\dfrac32$
$\to 2m\le-\dfrac32$
$\to m\le-\dfrac34$
