Question

39.7.
1) при x=77
$\frac{ {x}^{2} - 19 }{x - 3} + \frac{10}{ x - 3}$
2)при y=25,1
$\frac{y + 7}{ {y}^{2} - 25 } - \frac{2y + 2}{ {y}^{2} - 25 }$
3)при x=-10.25
$\frac{ {x}^{2} - 43 }{x - 6} - \frac{7}{6 - x}$
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Answer 1

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

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