Question

Giải các phương trình sau

a) \(2x^4-x^3-6x^2-x+2=0\)

b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

Answer 1

a: \(\Leftrightarrow2x^4-4x^3+3x^3-6x^2-x+2=0\)

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

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

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

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

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

b: \(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)

\(\Leftrightarrow\left(x^2+5x\right)\cdot\left(x^2+5x+10\right)=0\)

=>x(x+5)=0

=>x=0 hoặc x=-5