Matemática, perguntado por anacaroline9134, 10 meses atrás

gente me ajuda por favor pelo amor de Deus​

Anexos:

Soluções para a tarefa

Respondido por dougOcara
0

Resposta:

\displaystyle\\h)\\\frac{2}{m^{2}n}-\frac{1}{mn^{2}}\\\\mmc(m^{2}n,mn^{2})\\m^{2}n,mn^{2}|mn\\m,n|m\\1,n|n\\1,1|=mn.m.n=m^{2}n^{2}\\\\\frac{2}{m^{2}n}-\frac{1}{mn^{2}}=\frac{2n-m}{m^{2}n^{2}} \\\\\\i)\\\frac{a-1}{4a}-\frac{a-3}{9a}\\\\mmc(4a,9a)\\4a,9a|a\\4,9|4\\1,9|9\\1,1|=4.9.a=36a\\\\\frac{a-1}{4a}-\frac{a-3}{9a}=\frac{9(a-1)-4(a-3)}{36a}=\frac{9a-9-4a+12)}{36a}=\frac{5a+3}{36a}\\\\j)\\\\\frac{2y-1}{7y}-\frac{2y-1}{14y}\\\\mmc(7y,14y)=\\7y,14y|y\\7,14|2\\7,7|7\\1,1|y.2.7=14y

\displaystyle\frac{2y-1}{7y}-\frac{2y-1}{14y}=\frac{2(2y-1)-(2y-1)}{14y} =\frac{2y-1}{14y}

Perguntas interessantes