Bài làm:
Ta có: \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}\)
\(=\frac{4-x^2}{x-3}+\frac{2x^2-2x}{x-3}+\frac{5-4x}{x-3}\)
\(=\frac{x^2-6x+9}{x-3}\)
\(=\frac{\left(x-3\right)^2}{\left(x-3\right)}=x-3\) \(\left(x\ne3\right)\)
\(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}.\)
\(=\frac{4-x^2}{x-3}-\frac{2x-2x^2}{x-3}+\frac{5-4x}{x-3}.\)
\(=\frac{4-x^2-2x+2x^2+5-4x}{x-3}\)
\(=\frac{x^2-6x+9}{x-3}\)
\(=\frac{\left(x-3\right)^2}{x-3}=x-3\)
\(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}\)ĐK : x \(\ne\)3
\(=\frac{4-x^2}{x-3}-\frac{2x-2x^2}{x-3}+\frac{5-4x}{x-3}\)
\(=\frac{4-x^2-2x+2x^2+5-4x}{x-3}=\frac{9+x^2-6x}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)