Question

алгебра 7 класс номер 6.50
ПОЖАЛУЙСЬА ПОМОГИТЕ​

Answer 1

Ответ:

Раскладываем знаменатели на множители .Применяем формулы сокращённого умножения .

$\bf \displaystyle 1)\ \ \frac{2m}{5m+5n}+\frac{3n}{5m-5n}=\frac{2m(5m-5n)+3n(5m+5n)}{5(m+n)\cdot 5(m-n)}=\\\\\\=\frac{10m^2+5mn+15n^2}{25(m^2-n^2)}=\frac{2m^2+mn+3n^2}{5(m^2-n^2)}\\\\\\2)\ \ \frac{7x}{3x+3y}-\frac{2x}{3x-3y}=\frac{7x(3x-3y)-2x(3x+3y)}{3(x+y)\cdot 3(x-y)}=\\\\\\=\frac{15x^2-27xy}{9(x^2-y^2)}=\frac{3x\cdot (5x-9y)}{9(x^2-y^2)}=\frac{x\cdot (5x-9y)}{3(x^2-y^2)}$          

$\bf \displaystyle 3)\ \ \frac{5b}{ax+ay}-\frac{2a}{bx+by}=\frac{5b}{a(x+y)}-\frac{2a}{b(x+y)}=\frac{5b^2-2a^2}{ab(x+y)}\\\\\\4)\ \ \frac{3x}{4x+4y}-\frac{6x}{8x+8y}=\frac{3x}{4(x+y)}-\frac{6x}{8(x+y)}=\frac{3x}{4(x+y)}-\frac{3x}{4(x+y)}=0$

Answer 2

$\displaystyle\bf\\1)\\\\\frac{2m}{5m+5n} +\frac{3n}{5m-5n} =\frac{2m}{5(m+n)} +\frac{3n}{5(m-n)} =\\\\\\=\frac{2m\cdot(m-n)+3n\cdot(m+n)}{5(m-n)(m+n)} =\frac{2m^{2} -2mn+3mn+3n^{2} }{5(m-n)(m+n)} =\\\\\\=\frac{2m^{2} +mn+3n^{2} }{5(m^{2} -n^{2} )}\\\\2)\\\\\frac{7x}{3x+3y} -\frac{2x}{3x-3y} =\frac{7x}{3(x+y)} -\frac{2x}{3(x-y)} =\\\\\\=\frac{7x\cdot(x-y)-2x\cdot(x+y)}{3(x-y)(x+y)} =\frac{7x^{2} -7xy-2xy-2x^{2} }{3(x-y)(x+y)} =\\\\\\=\frac{5x^{2}-9xy }{3(x^{2} -y^{2} )}$

$\displaystyle\bf\\3)\\\\\frac{5b}{ax+ay} -\frac{2a}{bx+by} =\frac{5b}{a(x+y)}-\frac{2a}{b(x+y)} =\\\\\\=\frac{5b\cdot b-2a\cdot a}{ab(x+y)}=\frac{5b^{2}-2a^{2} }{ab(x+y)} \\\\4)\\\\\frac{3x}{4x+4y} -\frac{6x}{8x+8y} =\frac{3x}{4(x+y)} -\frac{6x}{8(x+y)} =\\\\\\=\frac{3x\cdot 2-6x}{8(x+y)} =\frac{6x-6x}{8(x+y)} =0$