Câu 4: Không có nghĩa khi x-3=0

=>x=3

Câu 5:

\(A=\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)

\(=\frac{\left(x^4-x^3\right)-\left(x-1\right)}{\left(x^4+x^3+x^2\right)+\left(2x^2+2x+2\right)}=\frac{x^3.\left(x-1\right)-\left(x-1\right)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}\)

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

\(\dfrac{x^3+3x^2-2}{x^3+3x+4}\)

\(=\dfrac{x^3+x^2+2x^2+2x-2x-2}{x^3-x+4x+4}\)

\(=\dfrac{\left(x+1\right)\left(x^2+2x-2\right)}{\left(x+1\right)\left(x^2-x+4\right)}\)

\(=\dfrac{x^2+2x-2}{x^2-x+4}\)

a) \(\dfrac{3x^2+6xy}{6x^2}=\dfrac{3x\left(x+2y\right)}{6x^2}=\dfrac{x+2y}{2x}\)

b) \(\dfrac{2x^2-x^3}{x^2-4}=\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x^2}{x+2}\)

c) \(=\dfrac{x+1}{x^3+1}=\dfrac{x+1}{\left(x+1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

`a, (3x^2+6xy)/(6x^2) = (x+2y)/(3x)`

`b, (2x^2-x^3)/(x^2-4) = (x^2(2-x))/((x-2)(x+2))`

`= -x^2/(x+2)`

`c, (x+1)/(x^3+1) = 1/(x^2-x+1)`

a kham khảo nha , e nhờ a e lm chứ ko phải e lm nha ! 

\(\left(x-2\right)\left(\frac{3}{x}+2-\frac{5}{2x}-4+\frac{8}{x^2}-4\right)\)

\(\left(x-2\right)\left[\left(\frac{3}{x}-\frac{5}{2x}\right)-6+\frac{8}{x^2}\right]\)

\(\left(x-2\right)\left(\frac{1}{2x}-6+\frac{8}{x^2}\right)\)

\(\left(x-2\right)\left(\frac{3}{x+2}-\frac{5}{2x-4}+\frac{8}{x^2-4}\right)\)

\(=\left(x-2\right)\left[\frac{3}{x+2}-\frac{5}{2\left(x-2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{3.2\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{8.2}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}-\frac{5\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\)

\(=\left(x-2\right)\left[\frac{6\left(x-2\right)-5\left(x+2\right)+16}{2\left(x-2\right)\left(x+2\right)}\right]\)

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

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

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

Đợi anh chút

\(\dfrac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}=\dfrac{y\left(x^2+2xy+y^2\right)}{2x^2+2xy-xy-y^2}=\dfrac{y\left(x+y\right)^2}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\dfrac{y\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{y\left(x+y\right)}{2x-y}\)