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: \(z \neq 10\)