`a)` ĐKXĐ là: `x \ne +-2`
`M=[3x-1]/[x^2-4]+4/[x-2]+[5x]/[x+2]`
`M=[3x-1+4(x+2)+5x(x-2)]/[(x-2)(x+2)]`
`M=[3x-1+4x+8+5x^2-10x]/[(x-2)(x+2)]`
`M=[5x^2-3x+7]/[x^2-4]`
__________________________________________________________
`b)` ĐKXĐ: `x \ne +-4`
`N=[4x+9]/[x^2-16]+[9x]/[x-4]+8/[x+4]`
`N=[4x+9+9x(x+4)+8(x-4)]/[(x-4)(x+4)]`
`N=[4x+9+9x^2+36x+8x-32]/[(x-4)(x+4)]`
`N=[9x^2+48x-23]/[x^2-16]`
a) ĐKXĐ: \(x\ne\pm2\)
\(M=\dfrac{3x-1}{x^2-4}+\dfrac{4}{x-2}+\dfrac{5x}{x+2}\\ =\dfrac{3x-1}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{3x-1+4x+8+5x^2-10x}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{5x^2-3x+7}{\left(x-2\right)\left(x+2\right)}\)
b) ĐKXĐ: \(x\ne\pm4\)
\(N=\dfrac{4x+9}{x^2-16}+\dfrac{9x}{x-4}+\dfrac{8}{x+4}\\ =\dfrac{4x+9}{\left(x-4\right)\left(x+4\right)}+\dfrac{9x\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\\ =\dfrac{4x+9+9x^2+36x+8x-32}{\left(x-4\right)\left(x+4\right)}\\ =\dfrac{9x^2+48x-23}{\left(x-4\right)\left(x+4\right)}\)
