Question

giải pt

a) \(\dfrac{2x^3-3x-10}{\left(2-x\right)^3}=-2\)

b) \(\left(x^2-2x\right)^2-2\left(x-1\right)^2-1=0\)

Answer 1

b: \(\Leftrightarrow\left(x^2-2x+1-1\right)^2-2\left(x-1\right)^2-1=0\)

\(\Leftrightarrow\left[\left(x-1\right)^2-1\right]^2-2\left(x-1\right)^2-1=0\)

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

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

hay \(x\in\left\{1;3;-1\right\}\)

a: \(\Leftrightarrow2x^3-3x-10=-2\left(8-12x+6x^2-x^3\right)\)

\(\Leftrightarrow2x^3-3x-10=-16+24x-12x^2+2x^3\)

\(\Leftrightarrow-3x-10+16-24x+12x^2=0\)

=>\(12x^2-27x+6=0\)

hay \(x\in\left\{2;\dfrac{1}{4}\right\}\)