• Предмет: Алгебра
  • Автор: mix289
  • Вопрос задан 2 года назад

буду благодарен если сможете помочь. алгебра 8 класс номера 3,4,5​

Приложения:

Ответы

Ответ дал: Аноним
1

3.

а) 1/6 × (2а+3b)

в) 2а²-ab+6a-6b/a²+6a

д) n²+m²/m²-n²

б) b³-a²/ab

г) 6x-2/x²-2x

e) 4a/a²-4

4.

а) 3х-7у/х²-у²

б) -2ху+16х/у²-4

в) 5/6х+18

г) 23/15х-30

д) 5а-2/а²-а

е) 6в-3/16в²

ж) 3/ав

Ж) 1/а²-2а

!!!!!!!

Ответ дал: Ляляляля109
1

3.

а)

\frac{a}{3} +\frac{b}{2} =\frac{2a}{6} +\frac{3b}{6} =\frac{2a+3b}{6}

б)

\frac{b^{2}}{a} -\frac{a}{b} =\frac{b^{3}}{ab} -\frac{a^{2}}{ab} =\frac{b^{3}-a^{2}}{ab}

в)

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

г)

\frac{x+3}{x-2} -\frac{x-1}{x} =\frac{x(x+3)}{x(x-2)} -\frac{(x-1)(x-2)}{x(x-2)} =\frac{x^{2}+3x}{x(x-2)} -\frac{x^{2}-3x+2}{x(x-2)} =\frac{x^{2}+3x-x^{2}+3x-2}{x(x-2)} =\frac{6x-2}{x^{2}-2x}

д)

\frac{n}{m+n} -\frac{m}{m-n} =\frac{n(m-n)}{(m-n)(m+n)} -\frac{m(m+n)}{(m-n)(m+n)} =\frac{mn-n^{2}-m^{2}-mn}{(m-n)(m+n)} =\frac{-(m^{2}+n^{2})}{m^{2}-n^{2}} =\\\\=-\frac{m^{2}+n^{2}}{m^{2}-n^{2}}

е)

\frac{a}{a+2} -\frac{a}{a-2} =\frac{a(a-2)}{(a-2)(a+2)} -\frac{a(a+2)}{(a-2)(a+2)} =\frac{a^{2}-2a-a^{2}-2a}{(a-2)(a+2)} =-\frac{4a}{a^{2}-4}

ж)

\frac{x^{2}-4x}{x^{2}-16} -\frac{2x}{x-4} =\frac{x^{2}-4x}{(x-4)(x+4)} -\frac{2x}{x-4} =\frac{x^{2}-4x}{(x-4)(x+4)} -\frac{2x(x+4)}{(x-4)(x+4)} =\frac{x^{2}-4x-2x^{2}-8x}{(x-4)(x+4)} =\\\\=\frac{-x^{2}-12x}{(x-4)(x+4)} =-\frac{x^{2}+12x}{x^{2}-16}

4.

а)

\frac{5}{x+y} -\frac{2}{x-y} =\frac{5(x-y)}{(x-y)(x+y)} -\frac{2(x+y)}{(x-y)(x+y)} =\frac{5x-5y-2x-2y}{(x-y)(x+y)} =\frac{3x-7y}{x^{2}-y^{2}}

б)

\frac{3x}{y-2} -\frac{5x}{y+2} =\frac{3x(y+2)}{y^{2}-4} -\frac{5x(y-2)}{y^{2}-4} =\frac{3xy+6x-5xy+10x}{y^{2}-4} =\frac{16x-2xy}{y^{2}-4}

в)

\frac{4}{3x+9} -\frac{1}{2x+6} =\frac{4}{3(x+3)} -\frac{1}{2(x+3)} =\frac{8}{6(x+3)} -\frac{3}{6(x+3)} =\frac{5}{6(x+3)} =\frac{5}{6x+18}

г)

\frac{6}{5x-10} +\frac{1}{3x-6} =\frac{6}{5(x-2)} +\frac{1}{3(x-2)} =\frac{18}{15(x-2)} +\frac{5}{15(x-2)} =\frac{23}{15x-30}

д)

\frac{2}{a} +\frac{3}{a-1} =\frac{2(a-1)}{a(a-1)} +\frac{3a}{a(a-1)}=\frac{2a-2+3a}{a(a-1)} =\frac{5a-2}{a^{2}-a}

е)

\frac{2b-3}{16b^{2}} +\frac{2}{8b} =\frac{2b-3}{16b^{2}} +\frac{4b}{16b^{2}} =\frac{2b-3+4b}{16b^{2}} =\frac{6b-3}{16b^{2}}

ж)

\frac{a+3}{ab} -\frac{1}{b} =\frac{a+3}{ab} -\frac{a}{ab} =\frac{a+3-a}{ab} =\frac{3}{ab}

5.

а)

\frac{3x-5}{x} -\frac{3y-2}{y} =\frac{y(3x-5)}{xy} -\frac{x(3y-2)}{xy} =\frac{3xy-5y-3xy+2x}{xy} =\frac{2x-5y}{xy}

б)

\frac{1}{a^{2}} +\frac{a-2}{a} =\frac{1}{a^{2}}+\frac{a(a-2)}{a^{2}} =\frac{1+a^{2}-2a}{a^{2}} =\frac{(1-a)^{2}}{a^{2}}

в)

\frac{b-a}{ab} -\frac{a-b}{b^{2}} =\frac{b(b-a)}{ab^{2}} -\frac{a(a-b)}{ab^{2}} =\frac{b^{2}-ab-a^{2}+ab}{ab^{2}} =\frac{b^{2}-a^{2}}{ab^{2}}

г)

\frac{a-1}{2(a-4)} +\frac{a}{a-4} =\frac{a-1+2a}{2(a-4)} =\frac{3a-1}{2a-8}

д)

\frac{x-1}{3(x-4)} -\frac{x-3}{2(x-4)} =\frac{2(x-1)}{6(x-4)} -\frac{3(x-3)}{6(x-4)} =\frac{2x-2-3x+9}{6(x-4)} =\frac{7-x}{6x-24}

е)

\frac{3y}{4(y-1)} +\frac{2y}{5(1-y)} =\frac{3y}{4(y-1)} -\frac{2y}{5(y-1)} =\frac{15y-8y}{20(y-1)} =\frac{7y}{20y-20}

ж)

\frac{2}{(a-2)(a+2)} -\frac{1}{a(a+2)} =\frac{2a}{a(a-2)(a+2)} -\frac{a-2}{a(a-2)(a+2)} =\frac{2a-a+2}{a(a-2)(a+2)} =\frac{a+2}{a(a-2)(a+2)} =\\\\=\frac{1}{a^{2}-2a}

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