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 

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

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

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

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

8x^2 - 26x + m 2x - 3 4x - 7 8x^2 - 12x -14x + m -14x + 21 m - 21

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

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

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

2n^2 - n + 2 2n + 1 n - 1 2n^2 + n -2n + 2 -2n - 1 3

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 : \(\frac{x-3}{x-2}=\frac{x-2-1}{x-2}=-\frac{1}{x-2}\)

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