Đáp án: $x=\dfrac43$
Giải thích các bước giải:
Ta có:
$(4x-1)\sqrt{x^2+1}=2x^2+2x+1$
$\to 2(4x-1)\sqrt{x^2+1}=4x^2+4x+2$
$\to (4x-1)\cdot 2\sqrt{x^2+1}-(4x-1)=4x^2+4x+2-(4x-1)$
$\to (4x-1)\cdot (2\sqrt{x^2+1}-1)=4x^2+3$
$\to (4x-1)\cdot (2\sqrt{x^2+1}-1)=4(x^2+1)-1$
$\to (4x-1)\cdot (2\sqrt{x^2+1}-1)=(2\sqrt{x^2+1})^2-1$
$\to (4x-1)\cdot (2\sqrt{x^2+1}-1)=(2\sqrt{x^2+1}-1)(2\sqrt{x^2+1}+1)$
Mà $2\sqrt{x^2+1}-1\ge 2\sqrt{0+1}-1>0$
$\to 4x-1=2\sqrt{x^2+1}+1$
$\to 2\sqrt{x^2+1}=4x-2$
$\to \sqrt{x^2+1}=2x-1$
$\to 2x-1\ge 0\to x\ge \dfrac12$
Khi đó $x^2+1=(2x-1)^2\to x=\dfrac43$ vì $x\ge\dfrac12$
