Giải thích các bước giải:
Bài 5:
a.$\sqrt{2x-1}+x^2-3x+1=0$
$\rightarrow \sqrt{2x-1}-x+x^2-(2x-1)=0$
$\rightarrow (\sqrt{2x-1}-x)+(x-\sqrt{2x-1})(x+\sqrt{2x-1})=0$
$\rightarrow(x-\sqrt{2x-1})(x+\sqrt{2x-1}-1)=0$
$\rightarrow x-\sqrt{2x-1}=0\rightarrow x^2=2x-1\rightarrow (x-1)^2=0\rightarrow x=1$
Hoặc $x+\sqrt{2x-1}-1=0$
$\rightarrow x-1=-\sqrt{2x-1}\rightarrow x\le 1$
$\rightarrow (x-1)^2=2x-1$
$\rightarrow x^2-4x+2=0$
$\rightarrow x=2-\sqrt{2}$ vì $x\ge \dfrac{1}{2}$