ข้อ 4 ละกัน
$(\sqrt{x+1}+\sqrt{x})^2-4(\sqrt{x+1}+\sqrt{x})+1 = 0$
$\sqrt{x+1}+\sqrt{x} = 2\pm\sqrt{3} = 2+\sqrt{3}$
$(\sqrt{x+1}-2)^2 = (\sqrt{3}-\sqrt{x})^2$
$x+1+4-4\sqrt{x+1} = x+3-2\sqrt{3}\sqrt{x}$
$1-2\sqrt{x+1} = -\sqrt{3}\sqrt{x}$
$1+4(x+1)-4\sqrt{x+1} = 3x$
$x+5 = 4\sqrt{x+1}$
$x^2+10x+25 = 16x+16$
$x^2-6x+9 = 0$
$x = 3$
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