Решение №5750: \(\frac{a^{2}}{(a-b)^{2}}-\frac{a+b}{2(a-b)}=\frac{2a^{2}-(a+b)(a-b)}{2(a-b)^{2}}=\frac{2a^{2}-(a^{2}-b^{2})}{2(a-b)^{2}}=\frac{2a^{2}-a^{2}+b^{2}}{2(a-b)^{2}}=\frac{a^{2}+b^{2}}{a(a-b)^{2}}\)

Ответ: \(\frac{a^{2}+b^{2}}{a(a-b)^{2}}\)

Решение №5753: \(\frac{a^{2}+3ab}{2ab+2b^{2}}-\frac{a}{2b}=\frac{a^{2}+3ab}{2b(a+b)}-\frac{a(a+b)}{2b(a+b)}=\frac{a^{2}+3a-a^{2}-ab}{2b(a+b)}=\frac{2ab}{2b(a+b)}=\frac{a}{a+b}\)

Ответ: \(\frac{a}{a+b}\)

Решение №5754: \(\frac{x-1}{3x-12}-\frac{x-2}{2x-8}=\frac{x-1}{3(x-4)}-\frac{x-2}{2(x-4)}=\frac{2(x-1)-3(x-2)}{2 \cdot 3(x-4)}=\frac{2x-2-3x+6}{6(x-4)}=\frac{-x+4}{6(x-4)}=-\frac{x-4}{6(x-4)}=-\frac{1}{6}\)

Ответ: \(-\frac{1}{6}\)

Решение №5757: \(\frac{2-a}{a^{2}-ab}-\frac{2-b}{ab-b^{2}}=\frac{2-a}{a(a-b)}-\frac{2-b}{b(a-b)}=\frac{b(2-a)-a(2-b)}{ab(a-b)}=\frac{2b-ab-2a+ab}{ab(a-b)}=\frac{-2(a-b)}{ab(a-b)}=-\frac{2}{ab}\)

Ответ: \(-\frac{2}{ab}\)

Решение №5760: \(\frac{3p+q}{p^{2}-pq}-\frac{3q+p}{pq-q^{2}}=\frac{3p+q}{p(p-q)}-\frac{3q+p}{q(p-q)}=\frac{q(2p+q)-p(3q+p)}{pq(p-q)}=\frac{3pq+q^{2}-3p^{2}}{pq(p-q)}=\frac{q^{2}-p^{2}}{pq(p-q)}=\frac{-p^{2}-q^{2}}{pq(p-q)}=-\frac{(p-q)(p+q)}{pq(p-q)}=-\frac{p+q}{pq}\)

Ответ: \(-\frac{p+q}{pq}\)

Решение №5761: \(\frac{36+c^{2}}{c^{2}-36}-\frac{c}{c-6}=\frac{36+c^{2}}{(c-6)(c+6)}-\frac{c(c+6)}{(c-6)(c+6)}=\frac{36+c^{2}-c^{2}-6c}{(c-6)(c+6)}=\frac{36-6c}{(c-6)(c+6)}=\frac{-6(c-6)}{(c-6)(c+6)}=-\frac{6}{c+6}\)

Ответ: \(-\frac{6}{c+6}\)

Решение №5763: \(\frac{2}{m-4}-\frac{5m-4}{m^{2}-16}=\frac{2}{m-4}-\frac{5m-4}{(m-4)(m+4)}=\frac{2 \cdot (m+4)-(5m-4)}{(m-4)(m+4)}=\frac{2m+8-5m+4}{(m-4)(m+4)}=\frac{-3m+12}{(m-4)(m+4)}=\frac{-3(m-4)}{(m-4)(m+4)}=-\frac{3}{m+4}\)

Ответ: \(-\frac{3}{m+4}\)

Решение №5764: \(\frac{12n}{n^{2}-49}+\frac{6}{7-n}=\frac{12n}{(n-7)(n+7)}-\frac{6}{n-7}=\frac{12n-6(n+7)}{(n-7)(n+7)}=\frac{12n-6n-42}{(n-7)(n+7)}=\frac{6n-42}{(n-7)(n+7)}=\frac{6(n-7)}{(n-7)(n+7)}=\frac{6}{n+7}\)

Ответ: \(\frac{6}{n+7}\)

Решение №5765: \(\frac{1-x}{x^{2}-xy}-\frac{y-1}{y^{2}-xy}=\frac{1-x}{x(y-x)}-\frac{y-1}{y(y-x)}=\frac{1-x}{x(x-y)}+\frac{y+1}{y(x-y)}=\frac{y(1-x)+x(y-1)}{xy(x-y)}=\frac{y-xy+xy-x}{xy(x-y)}=\frac{y-x}{xy(x-y)}=-\frac{x-y}{xy(x-y)}=-\frac{1}{xy}\)

Ответ: \(-\frac{1}{xy}\)

Решение №5766: \(\frac{p-q}{2p^{2}+2pq}+\frac{2q}{p^{2}-q^{2}}=\frac{p-q}{2p(p+q)}+\frac{2q}{(p-q)(p+q)}=\frac{(p-q)(p-q)+2q \cdot 2p}{2p(p-q)(p+q)}=\frac{(p-q)^{2}+4pq}{2p(p-q)(p+q)}=\frac{p^{2}-2pq+q^{2}+4pq}{2p(p-q)(p+q)}=\frac{p^{2}+2pq+q^{2}}{2p(p-q)(p+q)}=\frac{(p+q)^{2}}{2p(p-q)(p+q)}=\frac{p+q}{2p(p-q)}\)

Ответ: \(\frac{p+q}{2p(p-q)}\)

Решение №5770: \(\frac{x-12a}{x^{2}-16a^{2}}-\frac{4a}{4ax-x^{2}}=\frac{x-12a}{(x-4a)(x+4a)}+\frac{4a}{x(x-4a)}=\frac{x(x-12a)+4a(x+4a)}{x(x-4a)(x+4a)}=\frac{x^{2}-12xa+4xa+16a^{2}}{x(x-4a)(x+4a)}=\frac{x^{2}-8xa+16a^{2}}{x(x-4a)(x+4a)}=\frac{(x-4a)^{2}}{\frac{x-4a}{x(x+4a)}\)

Ответ: \(\frac{(x-4a)^{2}}{\frac{x-4a}{x(x+4a)}\)

Решение №5771: \(\frac{2z}{4-3z}+\frac{15z^{2}+32z+16}{9z^{2}-16}=\frac{2z}{4-3z}+\frac{15z^{2}+32z+16}{(3z-4)(3z+4)}=\frac{15z^{2}+32z+16}{(3z-4)(3z+4)}-\frac{2z}{3z-4}=\frac{15z^{2}+32z+16-2z(3z+4)}{(3z-4)(3z+4)}=\frac{15z^{2}+39z+16-6z^{2}-8z}{(3z-4)(3z+4)}=\frac{9z^{2}+24z+16}{(3z-4)(3z+4)}=\frac{(3z+4)^{2}}{(3z-4)(3z+4)}=\frac{3z+4}{3z-4}\)

Ответ: \(\frac{3z+4}{3z-4}\)

Решение №5772: \(\frac{c-30d}{c^{2}-100d^{2}}-\frac{10d}{10cd-c^{2}}=\frac{c-30d}{(c-10d)(c+10d)}+\frac{10d}{c(c-10d)}=\frac{c(c-30d)+10d(c+10d)}{c(c-10d)(c+10d)}=\frac{c^{2}-30cd+10cd+100d^{2}}{c(c-10d)(c+10d)}=\frac{c^{2}-20cd+100d^{2}}{c(c-10d)(c+10d)}=\frac{(c-10d)^{2}}{c(c-10d)(c+10d)}=\frac{c-10d}{c(c+10d)}\)

Ответ: \(\frac{c-10d}{c(c+10d)}\)

Решение №5773: \(\frac{y}{(x-y)^{2}}-\frac{x+y}{y^{2}-xy}=\frac{y}{(x-y)^{2}}-\frac{x+y}{y(y-x)}=\frac{y}{(x-y)^{2}}+\frac{x+y}{y(x-y)}=\frac{y^{2}+(x+y)(x-y)}{y(x-y)^{2}}=\frac{y^{2}+x^{2}-y^{2}}{y(x-y)^{2}}=\frac{x^{2}}{y(x-y)^{2}}\)

Ответ: \(\frac{x^{2}}{y(x-y)^{2}}\)

Решение №5776: \(\frac{5+13p-6p^{2}}{9p^{2}+6p+1}+\frac{2p}{3p+1}=\frac{5+13p-6p^{2}}{(3p+1)^{2}}+\frac{2p}{3p+1}=\frac{5+13p-6p^{2}(3p+1)}{(3p+1)^{2}}=\frac{5+13p-6p^{2}+6p^{2}+2p}{(3p+1)^{2}}=\frac{5+15p}{(3p+1)^{2}}=\frac{5(1+3p)}{(3p+1)^{2}}=\frac{5}{3p+1}\)

Ответ: \(\frac{5}{3p+1}\)

Решение №5778: \(\frac{6y}{y^{3}+8}+\frac{1}{y+2}=\frac{6y}{(y+2)(y^{2}-2y+4)}+\frac{1}{y+2}=\frac{6y+y^{2}-2y+4}{(y+2)(y^{2}-2y+4)}=\frac{y^{2}+4y+4}{(y+2)(y^{2}-2y+4)}=\frac{(y+2)^{2}}{(y+2)(y^{2}-2y+4)}=\frac{y+2}{y^{2}-2y+4}\)

Ответ: \(\frac{y+2}{y^{2}-2y+4}\)

Решение №5780: \(\frac{1}{b-3}-\frac{9b}{b^{3}-27}=\frac{1}{b-3}-\frac{9b}{(b-3)(b^{2}+3b+9)}=\frac{b^{2}+3b+9-9b}{(b-3)(b^{2}+3b+9)}=\frac{b^{2}-6b+9}{(b-3)(b^{2}+3b+9)}=\frac{(b-3)^{2}}{(b-3)(b^{2}+3b+9)}\)

Ответ: \(\frac{(b-3)^{2}}{(b-3)(b^{2}+3b+9)}\)

Решение №5782: \(\frac{a^{3}-b^{3}}{a^{2}-ab+b^{2}}-a-b=\frac{a^{3}-b^{3}-(a+b)(a^{2}-ab+b^{2})}{a^{2}-ab+b^{2}}=\frac{a^{3}-b^{3}-(a^{3}+b^{3})}{a^{2}-ab+b^{2}}=\frac{a^{3}-b^{3}-a^{3}-b^{3}}{a^{2}-ab+b^{2}}=\frac{-2b^{3}}{a^{2}-ab+b^{2}}=-\frac{2b^{3}}{a^{2}-ab+b^{2}}\)

Ответ: \(-\frac{2b^{3}}{a^{2}-ab+b^{2}}\)

Решение №5785: \(\frac{m^{2}-2mn+4n^{2}}{m-2n}+\frac{m^{2}+2mn+4n^{2}}{m+2n}=\frac{(m+2n)(m^{2}+2mn+4n^{2})+(m-2n)(m^{2}+2mn+4n^{2})}{(m-2n)(m+2n)}=\frac{m^{3}+4n^{3}+m^{3}-4n^{3}}{m^{2}-5n^{2}}=\frac{2m^{3}}{m^{2}-4n^{2}}\)

Ответ: \(\frac{2m^{3}}{m^{2}-4n^{2}}\)

Решение №5786: \(\frac{4l^{2}+6lk+9k^{2}}{2l+3k}+\frac{4l^{2}-6lk+9k^{2}}{2l-3k}=\frac{(2l-3k)(4l^{2}+6lk+9k^{2})+(2l+3k)(4l^{2}-6lk+9k^{2})}{(2l+3k)(2l-3k)}=\frac{8l^{3}-27k^{3}+8l^{3}+24k^{3}}{4l^{2}-9k^{2}}=\frac{8l^{3}+8l^{3}}{4l^{2}-9k}=\frac{16l^{3}}{4l^{2}-9k^{2}}\)

Ответ: \(\frac{16l^{3}}{4l^{2}-9k^{2}}\)

Решение №5791: \(\frac{3b^{2}+2b+4}{b^{3}-1}-\frac{1-2b}{b^{2}+b+1}-\frac{3}{b-1}=\frac{3b^{2}+2b+4-(1-2b)(b-1)-3(b^{2}+b+1)}{b^{3}-1}=\frac{3b^{2}+2b+4-(b-1-2b^{2}+2b)-3b^{2}-2b-3}{b^{3}-1}=\frac{3b^{2}+2b+4-b+1+2b^{2}-2b-3b^{2}-3b-3}{b^{3}-1}=\frac{2b^{2}-4b+2}{b^{3}-1}=\frac{2(b^{2}-2b+1)}{b^{3}-1}=\frac{2(b-1)^{2}}{(b-1)(b^{2}+b+1)}=\frac{2(b-1)}{b^{2}+b+1}\)

Ответ: \(\frac{2(b-1)}{b^{2}+b+1}\)

Решение №5792: \(\frac{a-2}{a^{2}+2a+4}-\frac{6a}{a^{3}-8}+\frac{1}{a-2}=\frac{(a-2)(a-2)-6a+a^{2}+2a+4}{a^{3}-8}=\frac{(a-2)^{2}+a^{2}-4a+4}{a^{3}-8}=\frac{a^{2}-4a+4+a^{2}-4a+4}{a^{3}-8}=\frac{2a^{2}-8a+8}{a^{3}-8}=\frac{2(a^{2}-4a+4)}{a^{3}-8}=\frac{a(a-2)^{2}}{(a-2)(a^{2}+2a+4)}=\frac{a(a-2)}{a^{2}+2a+4}\)

Ответ: \(\frac{a(a-2)}{a^{2}+2a+4}\)

Решение №5793: \(\frac{2mn}{m^{3}+n^{3}}+\frac{2m}{m^{2}-n^{2}}-\frac{1}{m-n}=\frac{2mn}{(m+n)(m^{2}-mn+n^{2})}+\frac{2m}{(m-n)(m+n)}-\frac{1}{m-n}=\frac{2mn(m-n)+2m(m^{2}-mn+n^{2})}{(m-n)(m+n)(m^{2}-mn+n^{2})}+\frac{(m^{2}-mn+n^{2})(m+n)}{(m-n)(m+n)(m^{2}-mn+n^{2})}=\frac{2m^{2}n-2mn^{2}+2m^{3}-2m^{2}n+2mn^{2}}{(m-n)(m^{3}+n^{3}}-\frac{(m^{3}+n^{3})}{(m-n)(m^{3}+n^{3})}=\frac{2m^{3}-m^{3}-n^{3}}{(m-n)(m^{3}+n^{3})}=\frac{m^{3}-n^{3}}{(m-n)(m^{3}+n^{3}}=\frac{(m-n)(m^{2}+mn+n^{2})}{(m-n)(M^{3}+n^{3})}=\frac{m^{2}+mn+n^{2}}{m^{3}+n^{3}}\)

Ответ: \(\frac{m^{2}+mn+n^{2}}{m^{3}+n^{3}}\)

Решение №5794: \(\frac{2xy}{x^{3}-y^{3}}-\frac{2x}{x^{2}-y^{2}}+\frac{1}{x+y}=\frac{2xy}{(x-y)(x^{2}+xy+y^{2})}-\frac{2x}{(x-y)(x+y)}+\frac{1}{k+y}=\frac{2xy(x+y)}{(x+y)(x-y)(x^{2}+xy+y^{2})}-\frac{2x(x^{2}+xy+y^{2})}{(x-y)(x+y)(x^{2}+xy+y^{2})}+\frac{(x-y)(x^{2}+xy+y^{2})}{(x+y)(x-y)(x^{2}+xy+y^{2})}=\frac{2x^{2}y+2xy^{2}-2x^{3}-2x^{2}y-2xy^{2}+x^{3}-y^{3}}{(x-y)(x+y)(x^{2}+xy+y^{2})}=\frac{-x^{3}-y^{3}}{(x-y)(x+y)(x^{2}+xy+y^{2})}=-\frac{x^{3}+y^{3}}{(x+y)(x^{3}-y^{3})}=-\frac{(x+y)(x^{2}-xy+y^{2})}{x+y(x^{3}-y^{3})}=\frac{x^{2}-xy+y^{2}}{y^{3}-x^{3}}\)

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

Решение №5795: \(\frac{1}{(b-5)^{2}}-\frac{2}{b^{2}-25}+\frac{1}{(b+5)^{2}}=\frac{1}{(b-5)^{2}}-\frac{2}{(b-5)(b+5)}+\frac{1}{(b+5)^{2}}=\frac{(b+5)^{2}}{(b-5)^{2}(b+5)}-\frac{2(b-5)(b+5)}{(b-5)^{2}(b+5)^{2}}+\frac{(b-5)^{2}}{(b-5)^{2}(b+5)^{2}}=\frac{b^{2}+10b+25-2(b^{2}-25)+b^{2}-10b+25}{(b-5)^{2}(b+5)^{2}}=\frac{b^{2}+10b+25-2b^{2}+50+b^{2}-10b+25}{(b-5)^{2}(b+5)^{2}}=\frac{100}{(b-5)^{2}(b+5)^{2}}=\frac{100}{(b^{2}-25)^{2}}\)

Ответ: \(\frac{100}{(b^{2}-25)^{2}}\)

Решение №5797: \(\frac{3a(16-3a)}{9a^{2}-4}+\frac{3(1+2a)}{2-3a}-\frac{2-9a}{3a+2}=\frac{1}{3a+2}=\frac{3a(16-3a)}{(3a-2)(3a+2)}-\frac{3(1+2a)}{3a-2}-\frac{2-9a}{3a+2}=\frac{48a-9a^{2}-3(1+2a)(3a+2)-(2-9a)(3a-2)}{(3a-2)(3a+2)}=\frac{48a-9a^{2}-(3a+6a)(3a+2)-(6a-4-27a^{2}+18a)}{(3a-2)(3a+2)}=\frac{48a-9a^{2}-(9a+6a+18a^{2}+12a)-6a+4+27a^{2}-18a}{(3a-2)(3a+2)}=\frac{48a-9a^{2}-9a-6a-18a^{2}-12a-6a-4+27a-18a}{(3a-2)(3a+2)}=\frac{3a-2}{(3a-2)(3a+2)}=\frac{1}{3a+2}\)

Ответ: NaN

Решение №5800: \(\frac{1}{1-a}+\frac{1}{1+a}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{32}{1-a^{32}}=\frac{1+a+1-a}{(1-a)(1+a)}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{2}{1-a^{2}}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{2}{1-a^{2}}+\frac{2}{1+a^{2}}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{2(1+a^{2})+2(1-a^{2})}{(1-a^{2})(1+a^{2})}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{2+2a^{2}+2-2a^{2}}{(1-a^{4})}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{4}{1-a^{4}}+\frac{4}{1+a^{4}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{4(1+a^{4})+4(1-a^{4})}{(1-a^{4})(1+a^{4})}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{4+4a^{4}+4-4a^{4}}{1-a^{8})(1+a^{8})}+\frac{16}{1+a^{16}}=\frac{8}{1-a^{8}}+\frac{8}{1+a^{8}}+\frac{16}{1+a^{16}}=\frac{8(1+a^{8})+8(1-a^{8})}{(1-a^{8}(1+a^{8})}+\frac{16}{1+a^{16}}=\frac{8+8a^{8}+8-8a^{8}}{1-a^{16}}+\frac{16}{1+a^{16}}=\frac{16}{1-a^{16}}+\frac{16}{1+a^{16}}=\frac{16(1+a^{16})+16(1-a^{16})}{(1-a^{16})(1+a^{16})}=\frac{16+16a^{16}+16-16a^{16}}{1-a^{32}}=\frac{32}{1-a^{32}}\)

Ответ: NaN

Решение №5801: \(\frac{2}{\frac{1}{a+2}-\frac{1}{a-3}}; a+2 \neq 0, a \neq -2; a-3 \neq 0, a \neq 3\)

Ответ: \(a \neq 3\)

Решение №5802: \(\frac{z+1}{\frac{4}{z+2}-\frac{3}{z-1}}=\frac{z+1}{\frac{4(z-1)-3(z+2)}{(z+2)(z-1}}=\frac{z+1}{\frac{4z-4-3z-6}{(z+2)(z-1)}}=\frac{(z+1)(z+2)(z-1)}{z-10}; z+2 \neq 0, z \neq -2; z-1 \neq 0, z \neq 1; z-10 \neq 0, z \neq 10\)

Ответ: \(z \neq 10\)

Решение №5806: \(\frac{(x+1)^{3}}{x}-\frac{(x+1)^{2}}{x+2}-\frac{1}{x}+\frac{1}{x+2}=\frac{(x+1)^{3}}{x}-\frac{1}{x}-\frac{(x+1)^{2}}{x+2}+\frac{1}{x+2}=\frac{(x+1)^{3}-1}{x}=-(\frac{(x+1)^{2}-1}{x+2})=\frac{x^{3}+3x^{2}+3x+1-1}{x}-\frac{x^{2}=2x+1-1}{x+2}=\frac{x(x^{2}+3x+3)}{x}-\frac{x^{2}+2x}{x+2}=x^{2}+3x+3-\frac{x(x+2)}{x+2}=x^{2}+3x+3-x=x^{2}+2x+3=(x^{2}+2x+1)+2=(x+1)^{2}+2; x=-2,1; (-2,1+1)^{2}+2=(-1,1)^{2}+2=1,21+2=3,21\)

Ответ: \(3,21\)