Question

дам 300 балоорв срочнооло​

Answer 1

Ответ:

$1)\ \ \displaystyle \frac{x^2-9}{2x^2+1}\cdot \Big(\frac{6x+1}{x-3}+\frac{6x-1}{x+3}\Big)=\\\\\\=\frac{(x-3)(x+3)}{2x^2+1}\cdot \frac{6x^2+18x+x+3+6x^2-18x-x+3}{(x-3)(x+3)}=\\\\\\=\frac{12x^2+6}{2x^2+1}=\frac{6\, (2x^2+1)}{2x^2+1}=6$

$\displaystyle 2)\ \ \Big(\frac{3}{25-c^2}+\frac{1}{c^2-10c+25}\Big)\cdot \frac{(5-c)^2}{2}+\frac{3c}{c+5}=\\\\\\=\Big(\frac{3}{(5-c)(5+c)}+\frac{1}{(5-c)^2}\Big)\cdot \frac{(5-c)^2}{2}+\frac{3c}{c+5}=\\\\\\=\frac{3\, (5-c)+(5+c)}{(5-c)^2(5+c)}\cdot \frac{(5-c)^2}{2}+\frac{3c}{c+5}=\frac{15-3c+5+c}{(5+c)\cdot 2}+\frac{3c}{5+c}=\\\\\\=\frac{20-2c+6c}{2\cdot (5+c)}=\frac{20+4c}{2\cdot (5+c)}=\frac{4\cdot (5+c)}{2\cdot (5+c)}=\frac{4}{2}=2$