Đặt f(x) = x³ + ax + b

      g(x) = x² + x - 2 = x² - x + 2x - 2 = x( x - 1 ) + 2( x - 1 ) = ( x - 1 )( x + 2 )

f(x) ⋮ g(x) <=> ( x³ + ax + b ) ⋮ ( x - 1 )( x + 2 )

<=> \(\hept{\begin{cases}\left(x^3+ax+b\right)\text{⋮}\left(x-1\right)\left[1\right]\\\left(x^3+ax+b\right)\text{⋮}\left(x+2\right)\left[2\right]\end{cases}}\)

Áp dụng định lí Bézout vào [1] :

f(x) ⋮ ( x - 1 ) <=> f(1) = 0

<=> 1 + a + b = 0

<=> a + b = -1 (1)

Áp dụng định lí Bézout vào [2] :

f(x) ⋮ ( x + 2 ) <=> f(-2) = 0

<=> -8 - 2a + b = 0

<=> -2a + b = 8 (2)

Từ (1) và (2) => \(\hept{\begin{cases}a+b=-1\\-2a+b=8\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-3\\b=2\end{cases}}\)( hpt lớp 9 mới học nên làm sơ sơ :33 )

Vậy a = -3 ; b = 2

P/s: Dùng hệ số bất định cũng được

:33 bt làm r nhwung vẫn k bruh

\(P=\frac{x}{x-1}+\frac{3x}{x+2}+\frac{x^3-5x^2+x}{x^2+x-2}\)

1,ĐKXĐ:\(x\ne1,x\ne-2\)

Rg:\(P=\frac{x}{x-1}+\frac{3x}{x+2}+\frac{x^3-5x^2+x}{x^2+x-2}\)

\(=\frac{x}{x-1}+\frac{3x}{x+2}+\frac{x^3-5x^2+x}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{3x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}+\frac{x^3-5x^2+x}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{x^2+2x+3x^2-3x+x^3-5x^2+x}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{x^3-x^2}{\left(x-1\right)\left(x+2\right)}=\frac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\frac{x^2}{x+2}\)

2.Tại \(x=\frac{1}{2}\)ta có:

\(\frac{\left(\frac{1}{2}\right)^2}{\frac{1}{2}+2}=\frac{1}{10}\) 

3.Ta có:\(\frac{x^2}{x+2}=\frac{x^2-4+4}{x+2}=\frac{x^2-4}{x+2}+\frac{4}{x+2}\)\(=x-2+\frac{4}{x+2}\)

Để \(x\in Z\Rightarrow x-2\in Z\Rightarrow\)Để \(P\in Z\)thì \(\frac{4}{x+2}\in Z\)

\(\Rightarrow x+2\inƯ\left(4\right)\)

\(\Rightarrow x+2\in\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x\in\left\{-1;-3;0;-4;2;-6\right\}\)(TMĐKXĐ)

Vậy với \(x\in\left\{-1;-3;0;-4;2;-6\right\}\)thì \(P\in Z\)

Diện tích hình thang cân ABCD ạ

\(C=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

\(D=\frac{2x-x^2}{x^2-4}=\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-x}{x+2}\)

Bài làm 

\(C=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

\(E=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-x}{x+2}\)

Bài làm 

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

\(A=\frac{2x+6}{\left(x-3\right)\left(x-2\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

\(B=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)