Question

Упростите номера 49,50 и 51
Даю 30 баллов

Answer 1

$\displaystyle\bf\\49)\\\\x^{3} -x^{2} y-xy^{2}+y^{3} =( x^{3} -x^{2} y)-(xy^{2}-y^{3}) = \\\\=x^{2} \cdot(x-y)-y^{2} \cdot(x-y)=(x-y)\cdot(x^{2} -y^{2})=\\\\=(x-y)\cdot(x-y)\cdot(x+y)=(x-y)^{2} \cdot(x+y)\\\\x=2,5 \ \ ; \ \ y=-1,5\\\\(2,5+1,5)^{2} \cdot(2,5-1,5)=4^{2} \cdot1=16$

$\displaystyle\bf\\50)\\\\1)\\\\m^{2} -\frac{2+m^{4} }{m^{2}-1 } =\frac{m^{2}\cdot(m^{2}-1)-(2+m^{4} ) }{m^{2}-1 }=\\\\\\=\frac{m^{4} -m^{2} -2-m^{4} }{m^{2}-1 } =-\frac{m^{2} +2}{m^{2}-1 } \\\\\\2)\\\\-\frac{m^{2} +2}{m^{2}-1 } :\frac{m^{2}+2 }{m-1} =-\frac{m^{2} +2}{(m-1)\cdot(m+1) } \cdot\frac{m-1}{m^{2}+2 } =\\\\\\=-\frac{1}{m+1}$

$\displaystyle\bf\\51)\\\\1)\\\\\frac{1}{(x-1)^{2} } -\frac{x}{1-x^{2} } =\frac{1}{(1-x)^{2} } -\frac{x}{(1-x)\cdot(1+x)} =\\\\\\=\frac{1+x-x\cdot(1-x)}{(1-x)^{2}\cdot(1+x) } =\frac{1+x-x+x^{2}}{(1-x)^{2}\cdot(1+x) } =\\\\\\=\frac{1+x^{2} }{(1-x)^{2}\cdot(1+x) } \\\\\\2)\\\\\frac{1-x^{2} }{1+x^{2} }\cdot\frac{1+x^{2} }{(1-x)^{2}\cdot(1+x) } =\frac{1}{1-x}$