Question

 

\(\text{(12x^2y^2- 6xy^2) : 3xy+2y}\)
\(\text{b. \dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}}\)\(\text{c. \dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} }\)

 

Answer 1

\(a,\left(12x^2y^2-6xy^2\right):3xy+2y=6xy^2\left(2x-1\right):3xy+2y=2y\left(2x-1\right)+2y=4xy-2y+2y=4xy\)

\(b,\dfrac{4}{x+1} + \dfrac{8}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{4\left(x-1\right)+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{x-1}\)

\(c,\dfrac{1 }{x+1}- \dfrac{1}{x-1} +\dfrac{ 2x}{x^2-1} \)

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

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

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

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

\(=\dfrac{2}{x+1}\)

 

Answer 2

\(a,=4xy-2y+2y=4xy\\ b,\dfrac{4}{x+1}+\dfrac{8}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x-4+8}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{4x+4}{\left(x+1\right)\left(x-1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\\ c,\dfrac{1}{x+1}-\dfrac{1}{x-1}+\dfrac{2x}{x^2-1}=\dfrac{x-1-x-1+2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2x-2}{\left(x-1\right)\left(x+1\right)}=\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+1}\)