![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x^3-4x^2-3x^2+6x+4x-8+a+8 chia hết cho x-2
=>a+8=0
=>a=-8
b: =>2x^3+x^2-x^2-0,5x-0,5x+0,25+m-0,25 chia hết cho 2x+1
=>m-0,25=0
=>m=0,25
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3x^3+2x^2-7x+a chia hêt cho 3x-1
=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1
=>a-2=0
=>a=2
c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4
=>3a+19=4
=>3a=-15
=>a=-5
d: 2x^3-x^2+ax+b chiahêt cho x^2-1
=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1
=>a+2=0 và b-1=0
=>a=-2 và b=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^3+x^2-4x+m⋮2x-1\)
\(\Leftrightarrow2x^3-x^2+2x^2-x-3x+\dfrac{3}{2}+m-\dfrac{3}{2}⋮2x-1\)
\(\Leftrightarrow m=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\left(x\right)⋮B\left(x\right)\)
\(\Leftrightarrow2x^3-x^2+2x^2-x-3x+\dfrac{3}{2}+m-\dfrac{3}{2}⋮2x-1\)
\(\Leftrightarrow m=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\dfrac{A\left(x\right)}{B\left(x\right)}=\dfrac{x^4-\dfrac{1}{2}x^3+\dfrac{1}{2}x^3-\dfrac{1}{4}x^2+\dfrac{9}{4}x^2-\dfrac{9}{8}x-\dfrac{15}{8}x+\dfrac{15}{16}+a-\dfrac{1}{16}}{2x-1}\)
Để A(x) chia hết cho B(x) thì a-1/16=0
hay a=1/16
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow2x^3+6x^2-x^2-3x+6x+18+m-13⋮x+3\)
hay m=13