Question

thực hiện các phép tính

a. \(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\)

b.\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\)

c.\(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}\)

Answer 1

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

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

\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}=\dfrac{x+6}{2\left(x+3\right)}=\dfrac{x+3+3}{2\left(x+3\right)}=\dfrac{1}{2}+\dfrac{3}{2\left(x+3\right)}\)

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

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

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

c ) \(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}\)

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

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

Answer 2

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

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

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

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

\(=\dfrac{x+6}{2x\left(x+3\right)}\)

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

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

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

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

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

c) \(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}\)

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

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

\(=\dfrac{1}{xy}.\)