Đáp án: $a-\dfrac1a=3$
Giải thích các bước giải:
ĐKXĐ: $a>0$
Đặt $\sqrt{a}=x, x>0$
$\to x^2-\dfrac1{x^2}= x+\dfrac1x$
$\to (x-\dfrac1x)(x+\dfrac1x)=x+\dfrac1x$
$\to (x-\dfrac1x)(x+\dfrac1x)-(x+\dfrac1x)=0$
$\to (x-\dfrac1x-1)(x+\dfrac1x)=0$
Vì $x>0$
$\to x-\dfrac1x-1=0$
$\to x-\dfrac1x=1$
$\to (x-\dfrac1x)^2=1$
$\to x^2+\dfrac1{x^2}-2=1$
$\to a-\dfrac1a-2=1$
$\to a-\dfrac1a=3$
