Đáp án+Giải thích các bước giải:

 a)

\frac{2}{x+3}+\frac{x}{3-x}-\frac{x-x^{2}}{x^{2}-9} (đk:x\ne\pm3)

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

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

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

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

=\frac{-2}{x-3}

=\frac{2}{3-x}

b)

\frac{2}{x-2}+\frac{3}{x+2}-\frac{18-5x}{x^{2}-4} (đk:x\ne\pm2)

=\frac{2.(x+2)}{(x-2).(x+2)}+\frac{3.(x-2)}{(x-2).(x+2)}-\frac{18-5x}{(x-2).(x+2)}

=\frac{2x+4+3x-6-18+5x}{(x-2).(x+2)}

=\frac{10x-20}{(x-2).(x+2)}

=\frac{10.(x-2)}{(x-2).(x+2)}

=\frac{10}{x+2}

c)

\frac{4}{x+2}+\frac{3}{2-x}+\frac{12}{x^{2}-4} (đk:x\ne\pm2)

=\frac{4.(x-2)}{(x-2).(x+2)}-\frac{3.(x+2)}{(x-2).(x+2)}+\frac{12}{(x-2).(x+2)}

=\frac{4x-8-3x-6+12}{(x-2).(x+2)}

=\frac{x-2}{(x-2).(x+2)}

=\frac{1}{x+2}

Đáp án: + Giải thích các bước giải:

a. 2/{x+3} + x/{3-x} - {x-x^2}/{x^2-9}  (x \ne ±3)

={2(x-3)}/{(x-3)(x+3)} - {x(x+3)}/{(x-3)(x+3)} - {x-x^2}/{x^2-9}

= {2x-6-x^2-3x-x+x^2}/{(x-3)(x+3)}

={-2x-6}/{(x-3)(x+3)}

={-2(x+3)}/{(x-3)(x+3)}

=2/{3-x}

b. 2/{x-2} + 3/{x+3} - {18-5x}/{x^2-4}  (x \ne ±2)

={2(x+2)}/{(x-2)(x+2)} + {3(x-2)}/{(x-2)(x+2)} - {18-5x}/{x^2-4}

= {2x+4+3x-6-18+5x}/{(x-2)(x+2)}

={10x-20}/{(x-2)(x+2)}

={10(x-2)}/{(x-2)(x+2)}

={10}/{x+2}

c. 4/{x+2} + 3/{2-x} + {12}/{x^2-4}  (x \ne ±2)

={4(x-2)}/{(x-2)(x+2)} - {3(x+2)}/{(x-2)(x+2)} + {12}/{x^2-4}

= {4x-8-3x-6+12}/{(x-2)(x+2)}

={x-2}/{(x-2)(x+2)}

=1/{x+2}