Question

пж помогите, срочно​

Answer 1

а)

$\frac{x}{x+1} +\frac{x}{x-1} =2\frac{2}{3} \\\\\frac{x(x-1)}{(x+1)(x-1)} +\frac{x(x+1)}{(x+1)(x-1)} =\frac{8}{3} \\\\\frac{x^{2}-x+x^{2}+x}{(x+1)(x-1)} =\frac{8}{3} \\\\\frac{2x^{2}}{x^{2}-1} =\frac{8}{3} \\\\6x^{2}=8x^{2}-8\\2x^{2}=8\\x^{2}=4\\x_{1}=2\\x_{2}=-2$

б)

$\frac{x}{x+4} +\frac{x}{x-4} =5\frac{5}{9} \\\\\frac{x(x-4)}{(x+4)(x-4)}+\frac{x(x+4)}{(x+4)(x-4)}=\frac{50}{9} \\\\ \frac{x^{2}-4x+x^{2}+4x}{x^{2}-16} =\frac{50}{9}\\\\\frac{2x^{2}}{x^{2}-16} =\frac{50}{9}\\\\18x^{2}=50x^{2}-800\\32x^{2} =800\\x^{2}=25\\x_{1}=5\\x_{2}=-5$

в)

$\frac{x+3}{x-3}+\frac{x-3}{x+3}=3\frac{1}{3}\\\\\frac{(x+3)^{2}}{(x+3)(x-3)} +\frac{(x-3)^{2}}{(x+3)(x-3)} =\frac{10}{3}\\\\\frac{x^{2}+6x+9+x^{2}-6x+9}{x^{2}-9} =\frac{10}{3} \\\\\frac{2x^{2}+18}{x^{2}-9}=\frac{10}{3}\\\\3(2x^{2}+18)=10x^{2}-90\\6x^{2}+54=10x^{2}-90\\4x^{2}=144\\x^{2}=36\\x_{1}=6\\x_{2}=-6$

г)

$\frac{5x+7}{x-2}-\frac{2x+21}{x+2} = 8\frac{2}{3} \\\\\frac{(5x+7)(x+2)}{(x-2)(x+2)} -\frac{(2x+21)(x-2)}{(x-2)(x+2)} =\frac{26}{3}\\\\\frac{5x^{2}+17x+14}{x^{2}-4} -\frac{2x^{2}+17x-42}{x^{2}-4} =\frac{26}{3}\\\\\frac{5x^{2}+17x+14-2x^{2}-17x+42}{x^{2}-4}= \frac{26}{3}\\\\\frac{3x^{2}+56}{x^{2}-4} =\frac{26}{3} \\\\9x^{2}+168=26x^{2}-104\\17x^{2}=272\\x^{2}=16\\x_{1}=4\\x_{2}=-4$