Решение №1910: \(\frac{m^{3}+n^{3}}{m-n}-m^{2}-mn-n^{2}=\frac{m^{3}+n^{3}-(m-n)(m^{2}+mn+n^{2})}{m-n}=\frac{m^{3}+n^{3}-(m^{3}-n^{3})}{m-n}=\frac{m^{3}+n^{3}-m^{3}+n^{3}}{m-n}=\frac{2n^{3}}{m-n}\)

Ответ: \(\frac{2n^{3}}{m-n}\)

Решение №1911: \(\frac{x^{3}+y^{3}}{x^{2}+xy+y^{2}}+x-y=\frac{x^{3}+y^{3}(x-y)(x^{2}+xy+y^{2})}{x^{2}+xy+y^{2}}=\frac{x^{3}+y^{3}+x^{3}-y^{3}}{x^{2}+xy+y^{2}}=\frac{2x^{3}}{x^{2}+xy+y^{2}}\)

Ответ: \(\frac{2x^{3}}{x^{2}+xy+y^{2}}\)

Решение №1914: \(1-\frac{1}{a^{3}+1}-\frac{a}{a+1}=1-\frac{1}{(a+1)(a^{2}-a+1)}-\frac{a}{a+1}=\frac{a^{3}+1-1-a(a^{2}-a+1)}{a^{3}+1}=\frac{a^{3}-a^{3}+a^{2}-a}{a^{3}+1}=\frac{a^{2}-a}{a^{3}+1}\)

Ответ: \(\frac{a^{2}-a}{a^{3}+1}\)

Решение №1917: \(\frac{1}{b+2}-\frac{b}{b^{2}-2b+4}-\frac{12}{b^{3}+8}=\frac{b^{2}-2b+4}{b^{3}+8}-\frac{b(b+2)}{b^{3}+8}-\frac{12}{b^{3}+8}=\frac{b^{2}-2b+4-b^{2}-2b-12}{b^{3}+8}=\frac{-4b-8}{b^{3}+8}=\frac{-4(b+2)}{(b+2)(b^{2}-2b+4)}=-\frac{4}{b^{2}-2b+4}=\frac{4}{2b-b^{2}-4}\)

Ответ: \(\frac{4}{2b-b^{2}-4}\)

Решение №1925: \(\frac{x+2y}{x^{2}+2xy+y^{2}}-\frac{x-2y}{x^{2}-y^{2}}+\frac{2y^{2}}{(x+y)(x^{2}-y^{2})}=\frac{2y}{x^{2}-y^{2}}=\frac{x+2y}{(x+y)^{2}}-\frac{x-2y}{(x-y)(x+y)}+\frac{2y^{2}}{(x+y)(x-y)(x+y)}=\frac{(x+2y)(x-y)}{(x+y)^{2}(x-y)}-\frac{(x-2y)(x+y)}{(x+y)^{2}(x-y)}+\frac{2y^{2}}{(x+y)^{2}(x-y)}=\frac{x^{2}-xy+2xy-2y^{2}-(x^{2}+xy-2xy-2y^{2})+2y^{2}}{(x+y)^{2}(x-y)}=\frac{2xy+2y^{2}}{(x+y)^{2}(x-y)}=\frac{2y(x+y)}{(x+y)^{2}(x-y)}=\frac{2y}{(x+y)(x-y)}=\frac{2y}{x^{2}-y^{2}}\)

Ответ: NaN

Решение №1931: \(\frac{x-2y}{x-\frac{x^{2}+y^{2}}{x-y}}=\frac{x-2y}{\frac{x(x-y)-x^{2}-y^{2}}{x-y}}=\frac{(x-2y)(x-y)}{x^{2}-xy-x^{2}-y^{2}}=\frac{(x-2y)(x-y)}{-xy-y}=-\frac{(x-2y)(x-y)}{y(x+y)}; x-y \neq 0, x \neq y, y \neq 0, x+y \neq 0, x \neq -y\)

Ответ: \(x \neq -y\)

Решение №1932: \(\frac{8}{(x^{2}+3}+\frac{2}{x^{2}+3}+\frac{1}{x+1}=\frac{8}{(x^{2}+3)(x=1)(x+1)}+\frac{2(x^{2}-1)}{(x^{2}+3)(x^{2}-1)}+\frac{(x^{2}+3)(x-1)}{(x^{2}+3)(x^{2}-1)}=\frac{8+2x^{2}-2+x^{3}-x^{2}+3x-3}{(x^{2}+3)(x^{2}-1)}=\frac{x^{3}+x^{2}+3x+3}{(x^{2}+3)(x^{2}-1)}=\frac{x^{2}(x+1)+3(x+1)}{(x^{2}+3)(x^{2}-1)}=\frac{(x^{2}+3)(x+1)}{(x^{2}+3)(x-1)(x+1)}=\frac{1}{x-1}; x=0,992; \frac{1}{0,992-1}=\frac{1}{-0,008}=-\frac{1000}{8}=-125\)

Ответ: \(-125\)

Решение №1941: \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+ … +\frac{1}{99 \cdot 100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+…+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}-\frac{100}{100}-\frac{1}{100}=\frac{99}{100}=0,99\)

Ответ: \(0,99\)

Решение №1943: \(\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 6}+ … +\frac{1}{99 \cdot 101}=\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+…+\frac{1}{2}(\frac{1}{99}-\frac{1}{100})=\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101})=\frac{1}{2}(1-\frac{1}{101})=\frac{1}{2} \cdot (\frac{101}{101}-\frac{1}{101})=\frac{1}{2} \cdot \frac{100}{101}=\frac{50}{101}\)

Ответ: \(\frac{50}{101}\)

Решение №1947: \(\frac{5c}{2d}:(-\frac{15c}{d})=frac{5c}{2d} \cdot (-\frac{d}{15c})=-\frac{5c \cdot d}{2d \cdot 15c}=-\frac{5}{30}=-\frac{1}{6}\)

Ответ: \(-\frac{1}{6}\)