Question

6.26. Докажите тождество: 2) (x - xy + 2 - zy)/(1 - 3y + 3y ^ 2 - y ^ 3) = (x + x)/((1 - y) ^ 1) 1) (ac + bx + ax + bc)/(ay + 2bx + 2ax + by) = (x + c)/(2x + y)​ "" дам 100 баллов срочно ""

Answer 1

$\displaystyle\bf\\1)\\\\\frac{ac+bx+ax+bc}{ay+2bx+2ax+by} =\frac{(ac+bc)+(ax+bx)}{(ay+by)+(2ax+2bx)} =\frac{c(a+b)+x(a+b)}{y(a+b)+2x(a+b)} =\\\\\\=\frac{(a+b)(x+c)}{(a+b)(2x+y)} =\frac{x+c}{2x+y} \\\\\\\frac{x+c}{2x+y} =\frac{x+c}{2x+y} \\\\\\2)\\\\\frac{x-xy+z-zy}{1-3y+3y^{2} -y^{3} } =\frac{(x-xy)+(z-zy)}{(1-y^{3})-(3y-3y^{2} ) } =\frac{x(1-y)+z(1-y)}{(1-y)(1+y+y^{2} )-3y(1-y)}=\\\\\\=\frac{(1-y)(x+z)}{(1-y)(1+y+y^{2} -3y)} =\frac{x+z}{1-2y+y^{2} } =\frac{x+z}{(1-y)^{2} }$

$\displaystyle\bf\\\frac{x+z}{(1-y)^{2} } =\frac{x+z}{(1-y)^{2} } \\\\\\3)\\\\\frac{3a^{3}+ab^{2} -6a^{2} b-2b^{3} }{9a^{5} -ab^{4} -18a^{4} b+2b^{5} } =\frac{(3a^{3} -6a^{2} b)+(ab^{2} -2b^{3} )}{(9a^{5}-18a^{4} b)-(ab^{4} -2b^{5} ) } =\\\\\\=\frac{3a^{2} (a-2b)+b^{2} (a-2b)}{9a^{4} (a-2b)-b^{4} (a-2b)} =\frac{(a-2b)(3a^{2} +b^{2} )}{(a-2b)(9a^{4} -b^{4} )} =\\\\\\=\frac{3a^{2} +b^{2} }{(3a^{2} +b^{2} )(3a^{2} -b^{2} )} =\frac{1}{3a^{2} -b^{2} }$

$\displaystyle\bf\\\frac{1}{3a^{2} -b^{2} } =\frac{1}{3a^{2} -b^{2} }$