Question

[nguyễn thị thảo vy]

giải pt : x⁴+2x³+5x²+4x-12=0

Answer 1

[๖Fly༉Donutღღ]

x^4 + 2x^3 + 5x^2 + 4x-12 = 0 
<=> (x^4 - x^3) + (3x^3-3x^2) + (8x^2 - 8x) + (12x-12) = 0 
<=> (x-1).(x^3 + 3x^2 + 8x+12) = 0 
<=> (x-1).[(x^3+2x^2)+(x^2+2x)+(6x+12)] = 0 
<=>(x-1).(x+2).(x^2+x+6) = 0 
<=> x= 1 hoặc x = -2 

Chúc học tốt ( hên xui đó nha )

Answer 2

[Nguyễn Xuân Anh]

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

\(\text{Vì }x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}\left(\text{nên vô No}\right)\)