Question

Плиз помогите решить 9а, и 10б.

Answer 1

Ответ:

9. a) $x=-9$;

10. б) $x=1$

Объяснение:

9. а)

$\frac{6}{x^{2} -9} +\frac{2}{x+3} =\frac{3}{x-3} \\\\\frac{6}{x^{2} -9} +\frac{2}{x+3} =\frac{3}{x-3}, x\neq -3, x\neq 3\\\\\frac{6}{x^{2} -9} +\frac{2}{x+3} =\frac{3}{x-3}=0\\\\\frac{6}{(x-3)*(x+3)} +\frac{2}{x+3} -\frac{3}{x-3} =0\\\\\frac{6+2(x-3)-3(x+3)}{(x-3)*(x+3)} =0\\\\\frac{6+2x-6-3x-9}{(x-3)*(x+3)} =0\\\\\frac{-x-9}{(x-3)*(x+3)} =0\\\\-x-9=0\\-x=9\\x=-9, x\neq -3, x\neq 3$

Ответ: $x=-9$

10. б)

$\frac{3x-5}{x+1} -\frac{x+5}{x-3} =2\\\\\frac{3x-5}{x+1} -\frac{x+5}{x-3} =2, x\neq -1, x\neq 3\\\\\frac{(x-3)*(3x-5)-(x+1)*(x+5)}{(x+1)*(x-3)} =2\\\\\frac{3x^{2} -5x-9x+15-(x^{2} +5x+x+5)}{(x+1)*(x-3)} =2\\\\\frac{3x^{2} -5x-9x+15-(x^{2} +6x+5)}{(x+1)*(x-3)} =2\\\\\frac{3x^{2} -5x-9x+15-x^{2} -6x-5}{(x+1)*(x-3)} =2\\\\\frac{2x^{2} -20x+10}{(x+1)*(x-3)} =2\\\\2x^{2} -20x+10=2(x+1)*(x-3)\\\\2x^{2} -20x+10=(x+1)*(x-3)\\\\2x^{2} -20x+10=2x^{2} -6x+2x-6\\\\-20x+10=-6x+2x-6\\\\-20x+10=-4x-6$

$-20x+4x=-6-10\\\\-16x=-16\\\\x=1,x\neq -1,x\neq 3$

Ответ: $x=1$