Bài 6:Giải phương trình: a 3x(2x-sqrt (x^2+3))=2(1-x^4)

1. **Expand both sides:** <br /> * Left side: $6x^2 - 3x\sqrt{x^2 + 3}$<br /> * Right side: $2 - 2x^4$<br /><br />2. **Rearrange to isolate the radical:**<br /> * $3x\sqrt{x^2 + 3} = 6x^2 + 2x^4 - 2$<br /><br />3. **Square both sides:**<br /> * $9x^2 (x^2 + 3) = (6x^2 + 2x^4 - 2)^2$<br /><br />4. **Simplify and rearrange into a polynomial equation:**<br /> * $2x^8 - 20x^6 + 27x^4 + 36x^2 - 4 = 0$<br /><br />5. **Notice that this equation is a quadratic in terms of x²:**<br /> * Let y = x²<br /> * The equation becomes: $2y^4 - 20y^3 + 27y^2 + 36y - 4 = 0$<br /><br />6. **Solve for y using factoring or numerical methods (e.g., the Rational Root Theorem or a graphing calculator).**<br /> * One solution is y = 1.<br /><br />7. **Substitute back x² for y and solve for x:**<br /> * x² = 1<br /> * x = 1 or x = -1<br /><br />8. **Check for extraneous solutions by plugging the solutions back into the original equation.**<br /> * x = 1 is a valid solution.<br /> * x = -1 is an extraneous solution.<br /><br />Therefore, the solution to the equation is **x = 1**. <br />