Буду благодарен если сможете алгебра 8 класс номера 3,4,5

Ответ:

oleg34643

18.11.2020 14:31

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)

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

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)

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

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)

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