Question

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

b)\((2x)^{3}-(x)^{2}+x+1:2x+1 \)

Answer 1

a) \(\dfrac{x+2}{x^2-4}-\dfrac{2}{x^2-2x}\left(đkxđ:x\ne2;x\ne-2;x\ne0\right).\)

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

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

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

Answer 2

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

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

 

Answer 3

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

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

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

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

b,k bt đề sai hay k mà giải k đc