Question

tìm nghiệm của phương trình sau:

\(\dfrac{16}{x^2-4}+\dfrac{x+2}{2-x}=\dfrac{2-x}{x+2}\)

Answer 1

\(\dfrac{16}{x^2-4}+\dfrac{x+2}{2-x}=\dfrac{2-x}{x+2}\) (ĐK: \(x\ge0;x\ne\pm2\))

\(\Leftrightarrow\dfrac{16}{x^2-2^2}+\dfrac{x+2}{2-x}=\dfrac{2-x}{x+2}\)

\(\Leftrightarrow\dfrac{16}{\left(x-2\right)\left(x+2\right)}+\dfrac{x+2}{2-x}=\dfrac{2-x}{x+2}\)

\(\Rightarrow16\left(2-x\right)+\left(x+2\right)^2\left(x-2\right)=\left(2-x\right)^2\left(x-2\right)\)

\(\Leftrightarrow24-20x+x^3+2x^2=12x-8-6x^2+x^3\)

\(\Leftrightarrow24-20x+2x^2=12x-8-6x^2\)

\(\Leftrightarrow24-20x+2x^2+12x+8+6x^2=0\)

\(\Leftrightarrow8x^2-32x+32=0\)

\(\Leftrightarrow x=2\) (loại)

PT vô nghiệm