PT $\Leftrightarrow 1+2(3x+2)\sqrt{2x^2-1}=10x^2+8x$ $\Leftrightarrow (9x^2+12x+4)-2(3x+2)\sqrt{2x^2-1}+2x^2-1=x^2+4x+4$ $\Leftrightarrow (3x+2-\sqrt{2x^2-1})^2=(x+2)^2$Đến đây xét 2TH là ok r!
PT $\Leftrightarrow 1+2(3x+2)\sqrt{2x^2-1}=10x^2+8x$ $\Leftrightarrow (9x^2+12x+4)-2(3x+2)\sqrt{2x^2-1}+2x^2-1=x^2+4x+4$ $\Leftrightarrow (3x+2-\sqrt{2x^2-1})^2=(x+2)^2$$\rightarrow TH1:3x+2-\sqrt{2x^2-1}=x+2\Leftrightarrow 2x=\sqrt{2x^2-1}\Leftrightarrow $
PT $\Leftrightarrow 1+2(3x+2)\sqrt{2x^2-1}=10x^2+8x$ $\Leftrightarrow (9x^2+12x+4)-2(3x+2)\sqrt{2x^2-1}+2x^2-1=x^2+4x+4$ $\Leftrightarrow (3x+2-\sqrt{2x^2-1})^2=(x+2)^2$
Đến đây x
ét
2
TH là o
k r!