Question

  giải giúp mình với ạ

Bài 1 (2 điểm). Thực hiện phép tính:

a) \(\frac{5x}{x-1} - \frac{x+1}{x-1}\)

b) \(\frac{4x+13}{5x(x-7)} - \frac{x-48}{5x(x-7)}\)

Bài 2 (2 điểm). Tính

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

Answer 1

Bài 1:

a: \(\dfrac{5x}{x-1}-\dfrac{x+1}{x-1}=\dfrac{5x-x-1}{x-1}=\dfrac{4x-1}{x-1}\)

b: \(\dfrac{4x+13}{5x\left(x-7\right)}-\dfrac{x-48}{5x\left(x-7\right)}\)

\(=\dfrac{4x+13-x+48}{5x\left(x-7\right)}=\dfrac{3x+61}{5x\left(x-7\right)}\)

Bài 2:

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

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

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

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