Question

39.7. 1) при x=772)при y=25,13)при x=-10.25​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{x^{2}-19 }{x-3} +\frac{10}{x-3} =\frac{x^{2} -19+10}{x-3} =\frac{x^{2} -9}{x-3} =\\\\\\=\frac{(x-3)(x+3)}{x-3} =x+3\\\\\\x=77 \ \ \ \Rightarrow \ \ \ x+3=77+3=80\\\\2)\\\\\frac{y+7}{y^{2} -25} -\frac{2y+2}{y^{2}-25 } =\frac{y+7-2y-2}{y^{2} -25} =-\frac{y-5}{y^{2} -25} =\\\\\\=-\frac{y-5}{(y-5)(y+5)} =-\frac{1}{y+5} \\\\\\y=25,1 \ \ \ \Rightarrow \ \ \ -\frac{1}{y+5} =-\frac{1}{25,1+5} =-\frac{1}{30,1} =-\frac{10}{301}$

$\displaystyle\bf\\3)\\\\\frac{x^{2} -43}{x-6} -\frac{7}{6-x} =\frac{x^{2} -43}{x-6} +\frac{7}{x-6} =\frac{x^{2} -43+7}{x-6}=\\\\\\=\frac{x^{2} -36}{x-6} =\frac{(x-6)(x+6)}{x-6} =x+6\\\\\\x=-10,25 \ \ \ \Rightarrow \ \ \ x+6=-10,25+6=-4,25$

Answer 2

It seems like you have provided three sets of values, and I'm assuming you would like me to perform some calculations using these values. However, you haven't specified what kind of calculations you would like me to perform. Could you please provide more information or clarify your question?