Question

Thực hiện phép tính
\(\frac{4-x^2}{x-3}\)\(+\)\(\frac{2x-2x^2}{3-x}\)\(+\)\(\frac{5-4x}{x-3}\)

Answer 1

Bài làm:

Ta có: \(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}\)

\(=\frac{4-x^2}{x-3}+\frac{2x^2-2x}{x-3}+\frac{5-4x}{x-3}\)

\(=\frac{x^2-6x+9}{x-3}\)

\(=\frac{\left(x-3\right)^2}{\left(x-3\right)}=x-3\) \(\left(x\ne3\right)\)

Answer 2

\(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}.\)

\(=\frac{4-x^2}{x-3}-\frac{2x-2x^2}{x-3}+\frac{5-4x}{x-3}.\)

\(=\frac{4-x^2-2x+2x^2+5-4x}{x-3}\)

\(=\frac{x^2-6x+9}{x-3}\)

\(=\frac{\left(x-3\right)^2}{x-3}=x-3\)

Answer 3

\(\frac{4-x^2}{x-3}+\frac{2x-2x^2}{3-x}+\frac{5-4x}{x-3}\)ĐK : x \(\ne\)3

\(=\frac{4-x^2}{x-3}-\frac{2x-2x^2}{x-3}+\frac{5-4x}{x-3}\)

\(=\frac{4-x^2-2x+2x^2+5-4x}{x-3}=\frac{9+x^2-6x}{x-3}=\frac{\left(x-3\right)^2}{x-3}=x-3\)