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

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 

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

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

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

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

Để \(A\left(x\right)⋮B\left(x\right)\)Khi mà chỉ khi : 

\(m-21=0\Leftrightarrow m=21\)

Vậy m = 21 thì \(A\left(x\right)⋮B\left(x\right)\)

Để 2 đa thức chia hết thì : \(2n+1\inƯ\left(3\right)=\left\{1;3\right\}\)

a + b , ĐKXĐ : \(x\ne2;-3\)

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

\(=\frac{x^2-4-5}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-3}{x-2}\)

c, Thay x = 2 ta có : ... Vì ko thỏa mãn giá trị của phân thức x khác 2 nên ko có giá trị biểu thức  

d, Ta có : \(\frac{x-3}{x-2}=\frac{x-2-1}{x-2}=-\frac{1}{x-2}\)

\(-x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)