Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)

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

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

\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

\(=\dfrac{x^2z}{3y}\)

\(\dfrac{6x-x^2-5}{5x^6-x^7}\\ =\dfrac{-x^2+5x+x-5}{x^6\left(5-x\right)}\\ =\dfrac{\left(-x^2+5x\right)+\left(x-5\right)}{x^6\left(5-x\right)}\\ =\dfrac{-x\left(x-5\right)+\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{\left(-x+1\right)\left(x-5\right)}{-x^6\left(x-5\right)}\\ =\dfrac{-x+1}{-x^6}\)

\(\dfrac{-x^2+6x-5}{5x^6-x^7}\)

\(=\dfrac{x^2-6x+5}{x^7-5x^6}\)

\(=\dfrac{\left(x-1\right)\left(x-5\right)}{x^6\left(x-5\right)}\)

\(=\dfrac{x-1}{x^6}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

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

\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

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

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

Làm chi tiết theo yêu cầu của bạn :

\(ĐKXĐ\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

\(\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)

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

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

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

\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

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

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

p/s : bonus nếu bạn không hiểu ptich đa thức thành nhân tử =))

x² + x - 6 = x² + 3x - 2x - 6 = x(x+3) - 2(x+3) = (x-2)(x+3)

x² - x - 12 = x² + 3x - 4x - 12 = x(x+3) - 4(x+3) = (x-4)(x+3)

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

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

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

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

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

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

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

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

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

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

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

b: \(B=\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

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

=x+y-z

c: 
ĐKXĐ: x<>1

\(C=\dfrac{x^2-6x+5}{x^2-2x+1}\)

\(=\dfrac{\left(x-1\right)\left(x-5\right)}{\left(x-1\right)^2}\)

\(=\dfrac{x-5}{x-1}\)

ko biết

\(2x^3\left(x^2-5\right)+\left(-2x^3+4x\right)+\left(6+x\right)x^2\)

\(=2x^5-10x^3-2x^3+4x+6x^2+x^3=2x^5-9x^3+6x^2+4x\)

ko biet