Question

Tính:

\(a,\dfrac{-5}{4+2y}+\dfrac{y-2}{2y+y^2}\)

\(b,\dfrac{x-1}{x^2-2xy}+\dfrac{3}{2xy-x^2}\)

Answer 1

a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)

\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+y\right)}\)

\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)

\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)

\(\frac{-3y-4}{2y\left(2+y\right)}\)

Answer 2

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

\(\frac{x-1}{x\left(x-2y\right)}+\frac{3}{x\left(2y-x\right)}\)

\(\frac{x-1}{x\left(x-2y\right)}+\frac{-3}{x\left(x-2y\right)}\)

\(\frac{x-1-3}{x\left(x-2y\right)}\)

\(\frac{x-4}{x\left(x-2\right)}\)

Answer 3

Nè bạn ơi, tớ không hiểu câu a của tớ bị làm sao lên tớ làm lại nhé

Answer 4

a)\(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}\)                                                             b)\(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}\)

\(\frac{-5}{2\left(2+y\right)}+\frac{y-2}{y\left(2+2\right)}\)                                                          \(\frac{x-1}{x^2-2xy}-\frac{3}{x^2-2xy}\)

\(\frac{-5y}{2y\left(2+y\right)}+\frac{2y-4}{2y\left(2+y\right)}\)                                                      \(\frac{x-1-3}{x^2-2xy}\)

\(\frac{-5y+2y-4}{2y\left(2+y\right)}\)                                                                          \(\frac{x-4}{x^2-2xy}\)

\(\frac{3y-4}{2y\left(2+y\right)}\)