Решение №11980: \(2s-\frac{(b+s)^{2}}{b}=\frac{2bs}{b}-\frac{b^{2}+2bs+s^{2}}{b}=\frac{2bs-b^{2}-2bs-s^{2}}{b}=\frac{-b^{2}-s^{2}}{b}=\frac{b^{2}+s^{2}}{b}\)

Ответ: \(\frac{b^{2}+s^{2}}{b}\)

Решение №11984: \(\frac{4}{3b}+3b+4=\frac{4}{36}+\frac{3b(3b+4)}{3b}=\frac{4+9b^{2}+12b}{3b}=\frac{(3b)^{2}+2 \cdot 2 \cdot 2 \cdot 3b+2^{2}}{3b}=\frac{(3b+2)^{2}}{3b}\)

Ответ: \(\frac{(3b+2)^{2}}{3b}\)

Решение №11985: \(\frac{9b^{2}-4}{3b}+(2-3b)=\frac{9b^{2}-4+6b-9b^{2}}{3b}=\frac{6b-4}{3b}\)

Ответ: \(\frac{6b-4}{3b}\)

Решение №11988: \(\frac{m+1}{m}-\frac{3m-1}{m^{2}}=\frac{m(m+1)}{m^{2}}-\frac{3m-1}{m^{2}}=\frac{m^{2}+m-3m+1}{m^{2}}=\frac{m^{2}-2m+1}{m^{2}}=\frac{(m-1)^{2}}{m^{2}}\)

Ответ: \(\frac{(m-1)^{2}}{m^{2}}\)

Решение №11990: \(\frac{5}{a}-\frac{10a-1}{5a^{3}}=\frac{5 \cdot 5a}{5a^{3}}-\frac{10a-1}{5a^{3}}=\frac{25a^{2}-10a+1}{5a^{3}}=\frac{(5a-1)^{2}}{5a^{3}}\)

Ответ: \(\frac{(5a-1)^{2}}{5a^{3}}\)

Решение №11992: \(12+\frac{4p}{q}+\frac{p^{2}}{3q^{2}}=\frac{12 \cdot 3q^{2}}{3q^{2}}+\frac{4p \cdot 3q}{3q^{2}}+\frac{p^{2}}{3q^{2}}=\frac{36q^{2}+12pq+p^{2}}{3q^{2}}=\frac{(6q+p)^{2}}{3q^{2}}\)

Ответ: \(\frac{(6q+p)^{2}}{3q^{2}}\)

Решение №11995: \(\frac{12x+5y}{4x^{2}y}-\frac{5y-4x}{5xy^{2}}=\frac{5y(12x+5y)}{20x^{2}y^{2}}-\frac{4x(5y-4x)}{20x^{2}y^{2}}=\frac{60xy+25y^{2}-20xy+16x^{2}}{20x^{2}y^{2}}=\frac{25y^{2}+40xy+16x6{2}}{20x^{2}y^{2}}=\frac{(5y+4x)^{2}}{20x^{2}y^{2}}; x=\frac{1}{2}, y=\frac{1}{5}; \frac{(5 \cdot \frac{1}{5} + 4 \cdot \frac{1}{2})^{2}}{20 \cdot (\frac{1}{2})^{2}(\frac{1}{5})^{2}}=\frac{(1+2)^{2}}{20 \cdot \frac{1}{4} \cdot \frac{1}{25}}=\frac{3^{2}}{5 \cdot \frac{1}{25}}=\frac{9}{\frac{1}{5}}=9 \cdot 5=45\)

Ответ: \(9 \cdot 5=45\)

Решение №12001: \(\frac{7n+2k}{9n-2k}+\frac{n}{2k}=\frac{2k(7n+2k)+n(9n-2k)}{2k(9n-2k)}=\frac{14nk+4k^{2}+9n^{2}-2nk}{2k(9n-2k)}=\frac{4k^{2}+12nk+9n^{2}}{2k(9n-2k)}=\frac{(2k+3n)}{2k(9n-2k)}; 9n-2k \neq 0, 9n \neq 2k, n \neq \frac{2k}{9}; 2k \neq 0, k \neq 0\)

Ответ: \(k \neq 0\)

Решение №12010: \(\frac{b}{a}+\frac{b}{a(a-1)}=\frac{b(a-1)+1}{a(a-1)}=\frac{ab-b+b}{a(a-1)}=\frac{ab}{a(a-1)}=\frac{b}{a-1}; a \neq 0, a-1 \neq 0; a \neq 1\)

Ответ: \(a \neq 1\)

Решение №12011: \(\frac{c+1}{c+3}-\frac{c^{2}-3}{c(c+3)}=\frac{c(c+1)-c^{2}+3}{c(c+3)}=\frac{c^{2}+c-c^{2}+3}{c(c+3)}=\frac{c+3}{c(c+3)}=\frac{1}{c}; c \neq 0, c+3 \neq 0; c \neq -3\)

Ответ: \(c \neq -3\)