Question

[Jimin]

tìm a,b:

a,\(x^4+ax^3+bx-1:x^2-1\)

b,\(x^4+x^3+ax^2+\left(a+b\right)x+2b+1:x^3+ax+b\)

Answer 1

[Mysterious Person]

đề thiếu nha . thiếu " phép chia này là phép chia hết " .

a) ta có : \(\dfrac{x^4+ax^3+bx-1}{x^2-1}\in Z\)

\(\Leftrightarrow x^4+ax^3+bx-1=\left(x^2-1\right)\left(x^2+cx+d\right)\)

\(\Leftrightarrow x^4+ax^3+bx-1=x^4+cx^3+dx^2-x^2-cx-d\)

\(\Leftrightarrow x^4+ax^3+bx-1=x^4+cx^3+\left(d-1\right)x^2-cx-d\)

\(\Rightarrow\left\{{}\begin{matrix}a=c\\d-1=0\\b=-c\\-1=-d\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}d=1\\a=c=-b\end{matrix}\right.\)

vậy ..............................................................

câu b lm tương tự