Question

2x^5-3x^4-5x^3+5x^2+3x-2=0

Answer 1

pt này là pt đối xứng bậc 5 để tui làm cho

pt<=> \(2x^5+2x^4-5x^4-5x^3+5x^2+5x-2x-2=0\)

<=> \(2x^4\left(x+1\right)-5x^3\left(x+1\right)+5x\left(x+1\right)-2\left(x+1\right)=0\)

<=> \(\left(2x^4-5x^3+5x-2\right)\left(x+1\right)=0\)

<=> \(\left[2x^4-2x^3-3x^3+3x^2-3x^2+3x+2x-2\right]\left(x+1\right)=0\)

<=> \(\left(x-2\right)\left(x+1\right)\left(2x^3-3x^2-3x+2\right)=0\)

<=> \(\left(x-2\right)^2\left(x+1\right)^2\left(2x-1\right)=0\)

<=>\(\hept{\begin{cases}x=-1\\x=2\\x=\frac{1}{2}\end{cases}}\)