x⁴ - 3x + 2 = (x⁴ - x³) + (x³ - x²) + (x² - x) + (-2x + 2)

= (x - 1)(x³ + x² + x - 2)

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

\(=x^4+ax^3+bx^2-2x-x^3-ax^2-bx+2\)

\(=x^4+\left(a-1\right)x^3+\left(b-a\right)x^2+\left(-b-2\right)x+2\)

Đồng nhất phần hệ số ;

\(\hept{\begin{cases}a-1=0\\b-a=0\\-b-2=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b-1=0\\b=1\end{cases}}\Leftrightarrow a=b=c\)

Vậy \(a=b=c=1\)

Cảm ơn  o0o Nguyễn Việt Hiếu o0o

Mình đã biết làm rồi

Thank you!

a: \(3x^3+a\cdot x^2+bx+9\)

\(=3x^3-27x+a\cdot x^2-9a+\left(b+27\right)x+9a+9\)

\(=\left(x^2-9\right)\left(3x+a\right)+x\left(b+27\right)+9a+9\)

Để \(3x^3+a\cdot x^2+b\cdot x+9\) chia hết cho \(x^2-9\) thì b+27=0 và 9a+9=0

=>a=-1 và b=-27

b: \(x^4+a\cdot x^3+bx-1\)

\(=x^4-x^2+a\cdot x^3-a\cdot x+x^2-1+\left(b+a\right)x\)

\(=\left(x^2-1\right)\left(x^2-a\cdot x+1\right)+\left(b+a\right)x\)

Để \(x^4+a\cdot x^3+bx-1\) chia hết cho \(x^2-1\) thì a+b=0

=>b=-a

bài 2:

\(A=\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)

\(=\left(c+b+a-2c\right)^3+\left(c+a+b-2b\right)^3\)

\(=\left(-2c\right)^3+\left(-2b\right)^3=-8\left(b+c\right)\)

sao nữa nhỉ :v

rồi sao nua

3x - 2 = 2x - 3

=> 3x - 2x = -3 + 2

=> 1x = -1

vậy x = -1

b, \(ax^3+bx^2+5x-50⋮\left(x^2+3x-10\right)\)

\(\Rightarrow f\left(x\right)=ax^3+bx^2+5x-50⋮\left(x-2\right)\left(x+5\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=0\\f\left(-5\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}f\left(2\right)=8a+4b+10-50=0\\f\left(-5\right)=-125a+25b-25-50=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=4\left(2a+b\right)=40\\f\left(-5\right)=-25\left(5a-b\right)=75\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}f\left(2\right)=2a+b=1\\f\left(-5\right)=5a-b=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{2}{7}\\b=\dfrac{11}{7}\end{matrix}\right.\)