Question

tìm x biết: \(\text{8x^3-12x^2+6x+1-(4x^2-1)=0}\)

Answer 1

Lời giải:

PT $\Leftrightarrow 8x^3-16x^2+6x+2=0$

$\Leftrightarrow (8x^3-8x^2)-(8x^2-8x)-(2x-2)=0$

$\Leftrightarrow 8x^2(x-1)-8x(x-1)-2(x-1)=0$

$\Leftrightarrow (x-1)(8x^2-8x-2)=0$

$\Leftrightarrow 2(x-1)(4x^2-4x-1)=0$

$\Leftrightarrow x-1=0$ hoặc $4x^2-4x-1=0$

Nếu $x-1=0\Leftrightarrow x=1$ 

Nếu $4x^2-4x-1=0$

$\Leftrightarrow (2x-1)^2-2=0$

$\Leftrightarrow (2x-1-\sqrt{2})(2x-1+\sqrt{2})=0$

$\Leftrightarrow x=\frac{1\pm \sqrt{2}}{2}$