Question

1.Giải các phương trình sau :
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
b)\(\left|x\right|-2\left|x-2\right|+3\left|x-3\right|=4\)
2. Tìm các số nguyên x;y thỏa mãn : \(x^3+3x=x^2y+2y+5\)

Answer 1

a)x⁵+2x⁴+3x³+3x²+2x+1=0

<=> x⁵+x⁴+x⁴+x³+2x³+2x²+x²+x+x+1=0

<=>x⁴(x+1)+x³(x+1)+2x²(x+1)+x(x+1)+(x+1)=0

<=>(x+1)(x⁴+x³+2x²+x+1)=0

<=>x²(x+1)(x²+x+2+\(\dfrac{1}{x^2}\))=0

<=>x²(x+1)[(x+\(\dfrac{1}{2}\))²+\(\dfrac{7}{4}+\dfrac{1}{x^2}\)]=0

Vì [(x+\(\dfrac{1}{2}\))²\(+\dfrac{7}{4}+\dfrac{1}{x^2}\)]>0 với mọi x thuộc R

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy S={0;-1}