Question

Tìm x

(8x^2-4x):(-4x)-(x+2)=8(2x^4-3x^3+x^2):(-1/2x^2)+4(x-1)²=0

Answer 1

\(\Leftrightarrow-2x+1-x-2=8\cdot\left(-4x^2+6x-2x\right)+4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow-3x-1+32x^2-48x+16x-4x^2+8x-4=0\)

\(\Leftrightarrow28x^2-27x-5=0\)

\(\text{Δ}=\left(-27\right)^2-4\cdot28\cdot\left(-5\right)=1289>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{27-\sqrt{1289}}{56}\\x_2=\dfrac{27+\sqrt{1289}}{56}\end{matrix}\right.\)