$4x\sqrt{3x^2{+1}}-\sqrt{x-1}=7x^{2}+1$
\Leftrightarrow $(4x\sqrt{3x^2+1}-\sqrt{x-1})^2=(7x^{2}+1)^2$
\Leftrightarrow $16x^2(3x^2+1)+x-1-2.4x\sqrt{3x^2+1}.\sqrt{x-1}=49x^4+14x^2+1$
\Leftrightarrow $48x^4+16x^2+x-1-8x\sqrt{(3x^2+1)(x-1)}=49x^4+14x^2+1$
\Leftrightarrow $49x^4+14x^2+1-48x^4-16x^2-x+1+8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $x^4 - 2x^2 - x + 2 + 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $x^4 - x^3 + x^3 - x^2 - x^2 + x - 2x + 2 + 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $x^3(x-1) + x^2(x-1) - x(x-1) - 2(x-1) + 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x-1)(x^3+x^2-x-2) + 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x-1)[x^2(x+1)-(x+1)+1] + 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x-1)[(x+1)(x^2-1)+1]+ 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x-1)(x+1)(x^2-1)+(x-1)+ 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x^2-1)(x^2-1)+(x-1)+ 8x\sqrt{(3x^2+1)(x-1)}=0$
\Leftrightarrow $(x^2-1)^2+(x-1)+8x\sqrt{(3x^2+1)(x-1)}=0$
Phương trình này chỉ có 1 nghiệm là $x=1$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
