Question

\((2x^3-x-1)^2-(x^2-7x+6)^2=0\)

Tìm x

Answer 1

x = 1

Answer 2

\(\left(2x^3-x-1\right)^2-\left(x^2-7x+6\right)^2=0\)

\(\Leftrightarrow\left(2x^3-x-1-x^2+7x-6\right)\left(2x^3-x-1+x^2-7x+6\right)=0\)

\(\Leftrightarrow\left(2x^3+6x-x^2-7\right)\left(2x^3-8x+x^2+5\right)=0\)

\(\Leftrightarrow x=1;x=-\dfrac{5}{2}\)

Answer 3

\(\Leftrightarrow\left(2x^3-x-1-x^2+7x-6\right)\left(2x^3-x-1+x^2-7x+6\right)=0\)

\(\Leftrightarrow\left(2x^3-x^2+6x-7\right)\left(2x^3+x^2-8x+5\right)=0\)

\(\Leftrightarrow\left(2x^3-2x^2+x^2-x+7x-7\right)\left(2x^3-2x^2+3x^2-3x-5x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(2x^2+x+7\right)\left(2x^2+3x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(2x^2+5x-2x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\cdot\left(2x+5\right)=0\)

hay \(x\in\left\{1;-\dfrac{5}{2}\right\}\)