Решение №11935: \(\frac{c^{2}+100}{c-10}+\frac{20c}{10-c}=\frac{c^{2}+100}{c-10}-\frac{20c}{c-10}=\frac{c^{2}-20c+100}{c-10}=\frac{(c-10)^{2}}{c-10}=c-10; c-10 \neq 0, c \neq 10\)

Ответ: \(c \neq 10\)

Решение №11936: \(\frac{x^{2}+y^{2}}{x^{2}-y^{2}}-\frac{2xy}{x^{2}-y^{2}}=\frac{x^{2}+y^{2}-2xy}{(x-y)(x+y)}=\frac{(x-y)^{2}}{(x-y)(x+y)}=\frac{x-y}{x+y}; x-y \neq 0, x \neq y; x+y \neq 0, x \neq -y\)

Ответ: \(x \neq -y\)

Решение №11937: \(\frac{d^{2}+49}{7-d}+\frac{14d}{d-7}=\frac{14d}{d-7}=\frac{d^{2}+49}{7-d}-\frac{14d}{7-d}=\frac{d^{2}+49-14d}{7-d}=\frac{(7-d)^{2}}{7-d}=7-d\)

Ответ: \(7-d\)

Решение №11939: \(\frac{n^{2}+n}{n^{3}-8}+\frac{n+4}{n^{3}-8}=\frac{n^{2}+n+n+4}{(n-2)(n^{2}+2n+4)}=\frac{n^{2}+2n+4}{(n-2)(n^{2}+2n+4)}=n-2; n-2 \neq 0, n \neq 2\)

Ответ: \(n \neq 2\)

Решение №11941: \(\frac{m^{2}+9}{m^{3}+27}-\frac{3m}{m^{3}+27}=\frac{m^{2}+9-3m}{m^{3}+27}=\frac{m^{2}-3m+9}{(m+3)(m^{2}-3m+9)}=\frac{1}{m+3}; m+3 \neq 0, m \neq -3\)

Ответ: \(m \neq -3\)

Решение №11942: \(\frac{b^{2}}{b^{2}+1}+\frac{2b^{2}+1}{b^{2}+1}-\frac{2(2b^{2}+1)}{b^{2}+1}=\frac{b^{2}+2b^{2}+1-4b^{2}-2}{b^{2}+1}=\frac{36^{2}-4b^{2}-1}{b^{2}+1}=\frac{-b^{2}-1}{b^{2}+1}=\frac{-(b^{2}+1)}{b^{2}+1}=-1\)

Ответ: NaN

Решение №11945: \(\frac{(m-1)^{2}}{m^{3}+27}+\frac{8-m}{m^{3}+27}=\frac{(m-1)^{2}+8-m}{m^{3}+27}=\frac{m^{2}-2m+1+8-m}{m^{3}+27}=\frac{m^{2}-3m+9}{(m+3)(m^{2}-3m+9)}=\frac{1}{m+3}m=-3,5; \frac{1}{m+3}=\frac{1}{-3,5+3}=\frac{1}{-0,5}=-\frac{1}{\frac{5}{10}}=-\frac{10}{5}=-2\)

Ответ: \(-2\)

Решение №11948: \(\frac{9x^{2}}{9x^{2}-4}-\frac{12x}{(3x-2)(3x+2)}+\frac{4}{9x^{2}-4}=\frac{9x^{2}}{9x^{2}-4}-\frac{12x}{9x^{2}-4}+\frac{4}{9x^{2}}=\frac{9x^{2}-12x+4}{9x^{2}-4}=\frac{(3x-2)^{2}}{(3x-2)(3x+2)}=\frac{3x-2}{3x+2}\)

Ответ: \(\frac{3x-2}{3x+2}\)

Решение №11951: \(\frac{8m^{2}+3m-2}{4m^{2}+4m+1}-\frac{5m-7}{-4m^{2}-4m-1}-\frac{4m-9}{(1+2m)^{2}}=\frac{8m^{2}+3m-2}{(1+2m)^{2}}+\frac{5m-7}{4m^{2}+4m+1}-\frac{4m-9}{(1+2m)^{2}}=\frac{8m^{2}+3m-2+5m-7-4m+9}{(1+2m)^{2}}=\frac{8m^{2}+4m}{(1+2m)^{2}}=\frac{4m(2m+1)}{(1+2m)^{2}}=\frac{4m}{1+2m}\)

Ответ: \(\frac{4m}{1+2m}\)

Решение №11953: \(\frac{2}{(3-a)(2-a)}+\frac{a-4}{(a-3)(a-2)}=\frac{2}{(3-a)(2-a)}+\frac{a-4}{(3-a)(2-a)}=\frac{2+a-4}{(3-a)(2-a)}=\frac{a-2}{(3-a)(2-a)}=\frac{a-2}{(a-3)(a-2)}=\frac{1}{a-3}\)

Ответ: \(\frac{1}{a-3}\)