Vì 3x^3+ax^2+bx+9 chia hết cho (x-3)(x+3) nên 3x^3+ax^2+bx+9  có dạng q(x)(x-3)(x+3)

f(3)=90+9a+3b=0 

f(-3)=-72+9a-3b=0

f(3)-f(-3)=162+6b=0

=> b=-27=> a=-1 

p/s mk lm hơi tắt mấy bước tính nhưng cách lm thì đầy đủ nha ~  

Ta có :

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

Đặt \(f_{\left(x\right)}=3x^3+ax^2+bx+9\)

Vì \(f_{\left(x\right)}⋮\left(x^2-9\right)\)

\(\Rightarrow\left\{{}\begin{matrix}f_{\left(3\right)}=3.3^3+a.3^2+3b+9=0\\f_{\left(-3\right)}=3.\left(-3\right)^3+a.\left(-3\right)^2+\left(-3\right)b+9=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}81+9a+3b+9=0\\-81+9a-3b+9=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}9a+3b+90=0\\9a-3b-72=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}9a+3b=-90\\9a-3b=72\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3a+b=-30\\3a-b=24\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6a=-6\\2b=-54\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-27\end{matrix}\right.\)

Vậy \(a=-1;b=-27\)

Cau a va b dat cot tim so du .Vi la phep chia het nen du bang 0.Cau c thi da thuc se chia het cho tich (x+3)(x-3) lam tuong tu hai cau a va b

Chào bạn!