Question

[Trang Nguyễn]

giải pt:

a) x⁴ + 2x³ -12x² -13x +42

b) (x -1) (x+2 )(x+4)(x+7)=16

c) x⁴ + 2x³ +5x² +4x -12 = 0

Answer 1

[Đời về cơ bản là buồn... cười!!!]

a) \(x^4+2x^3-12x^2-13x+42=0\)

\(\Leftrightarrow x^4+3x^3-x^3-3x^2-9x^2-27x+14x+42=0\)

\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)-9x\left(x+3\right)+14\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3-x^2-9x+14\right)=0\)

Answer 2

[Nhã Doanh]

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x^2+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)

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

Ta có:

\(x^2+x+6=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{23}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy...........