Question

Tìm \(x\) biết:

\(\left(2x-1\right)^3\)\(-2x\left(4x^2-3x\right)=5\)

Answer 1

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-6x^2\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+6x^2=5\)

\(\Leftrightarrow6x^2-6x+6=0\)

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

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0;\forall x\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0;\forall x\)

\(\Rightarrow\) Phương trình vô nghiệm

Answer 2

Ta có: \(\left(2x-1\right)^3-2x\left(4x^2-3x\right)=5\)

\(\Leftrightarrow8x^3-6x^2+12x-1-8x^3+6x^2=5\)

\(\Leftrightarrow12x=6\)

hay \(x=\dfrac{1}{2}\)