Toán 10( Đại số)

B

braga

$$(4x-1).\sqrt{x^2+1}=2x^2+2x+1$$
$$\Leftrightarrow (4x-1)\sqrt{x^2+1}=2(x^2+1)+2x-1$$
Đặt $\sqrt{x^2+1}=t(t>0)$ . (1) trở thành $$2t^2-(4x-1)t+2x-1=0$$
$\Delta = 16x^2-8x+1-8(2x-1)=(4x-3)^2$
 
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