Chủ đề này có 1 trả lời

[mrdoctorlee]

giải pt vô tỉ sau

$3\sqrt{6x^{2}-x-1}+\frac{46x+17}{\sqrt{2x-1}-4\sqrt{3x+1}}=5-8x$

[mrdoctorlee]

mk làm dcj cách này ko b t có sai sót j ko dk x>=1/2

$3\sqrt{6x^{2}-x-1}+\frac{(46x+17)(\sqrt{2x-1}+4\sqrt{3x+1})}{(\sqrt{2x-1}-4\sqrt{3x+1})(\sqrt{2x-1}+4\sqrt{3x+1})}=5-8x$

$\Leftrightarrow 3\sqrt{6x^{2}-x-1}-\sqrt{2x-1}-4\sqrt{3x+1}=5-8x$

$\Leftrightarrow 6(\sqrt{6x^{2}-x-1}-2)+2\sqrt{2x-1}(\sqrt{2x-1}-1)+4\sqrt{3x+1}(\sqrt{3x+1}-2)+12-2(2x-1)-4(3x+1)-10+16x=0$

$\Leftrightarrow \frac{6(6x+5)(x-1)}{\sqrt{6x^{2}-x-1}+2}+\frac{2\sqrt{2x-1}(2x-2)}{\sqrt{2x-1}+1}+\frac{4\sqrt{3x+1}(3x-3)}{\sqrt{3x+1}+2}=0$

$\Leftrightarrow (x-1)(\frac{6(6x+5)}{\sqrt{6x^{2}-x-1}+2}+\frac{4\sqrt{2x-1}}{\sqrt{2x-1}+1}+\frac{12\sqrt{3x+1}}{\sqrt{3x+1}+2})=0$

nên suy ra x=1