Question

RÚT GỌN PHÂN THỨC:

\(p=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

mk sửa lại câu hỏi ạ!

Answer 1

\(P=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{4-x^2}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\left(\dfrac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}-\dfrac{\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{\left(4+4x+x^2\right)+4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{4x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(\Leftrightarrow P=\dfrac{4x\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)

\(\Leftrightarrow P=\dfrac{4x\left(2+x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)\left(x-3\right)}\)

\(\Leftrightarrow P=\dfrac{4x}{x-3}\)

Answer 2

\(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(P=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(P=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{-\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(P=\left(\dfrac{2+x}{2-x}-\dfrac{-4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

\(P=\left(\dfrac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}-\dfrac{-4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{\left(2-x\right)\left(2-x\right)}{\left(2+x\right)\left(2-x\right)}\right):\dfrac{x-3}{2-x}\)

\(P=\left(\dfrac{\left(2+x\right)\left(2+x\right)+4x^2-\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x-3}{2-x}\)

\(P=\dfrac{4+2x+2x+x^2+4x^2-\left(4-2x-2x+x^2\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(P=\dfrac{4+2x+2x+x^2+4x^2-4+2x+2x-x^2}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(P=\dfrac{4x\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{2-x}\)

\(P=\dfrac{4x}{2-x}.\dfrac{2-x}{x-3}\)

\(P=\dfrac{4x}{x-3}\)

Answer 3

P=\(\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x-3}{2-x}\)

=\(\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{4-x^2}-\dfrac{2-x}{2+x}\right).\dfrac{2-x}{x-3}\)

=\(\left(\dfrac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right).\dfrac{2-x}{x-3}\)=\(\dfrac{4+4x+x^2+4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)=\(\dfrac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{2-x}{x-3}\)

=\(\dfrac{4x^2+8x}{\left(2-x\right)\left(x-2\right)}.\dfrac{2-x}{x-3}\)

=\(\dfrac{4x\left(x+2\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)\cdot\left(x-3\right)}\)

=\(\dfrac{4x}{x-3}\)