Verified answer

5)

[tex]=(\frac{m^2(m^2-1)-(2+m^4)}{m^2-1}):\frac{m^2+2}{m-1}=(\frac{m^4-m^2-2-m^4}{m^2-1})*\frac{m-1}{m^2+2}=\frac{-(m^2+2)}{(m-1)(m+1)}*\frac{m-1}{m^2+2}=-\frac{1}{m+1}[/tex]

D.

6)

[tex]\left \{ {{x^2-y^2+2x-4=0} \atop {x+y=0}} \right.[/tex]

x= -y

(-y)²-y²+2(-y)-4=0

x=-y

y²-y²-2y-4=0

-2y=4

y= -2

x= 2

(2;-2) C

[tex]\displaystyle\bf\\5)\\\\\Big(m^{2} -\frac{2+m^{4} }{m^{2} -1} \Big):\frac{m^{2} +2}{m-1} =\frac{m^{2} \cdot(m^{2} -1)-2-m^{4} }{(m-1)\cdot(m+1)} \cdot\frac{m-1}{m^{2}+2 } =\\\\\\=\frac{m^{4} -m^{2}-2-m^{4} }{(m-1)\cdot(m+1)}\cdot\frac{m-1}{m^{2} +2} =-\frac{ m^{2}+2 }{(m-1)\cdot(m+1)}\cdot\frac{m-1}{m^{2} +2} =\\\\\\=-\frac{1}{m+1} \\\\\\Otvet \ : \ D[/tex]

[tex]\displaystyle\bf\\6)\\\\\left \{ {{x^{2} -y^{2} +2x-4=0} \atop {x+y=0}} \right. \\\\\\\left \{ {{x^{2} -(-x)^{2} +2x-4=0} \atop {y=-x}} \right. \\\\\\\left \{ {{x^{2} -x^{2} +2x-4=0} \atop {y=-x}} \right\\\\\\\left \{ {{2x-4=0} \atop {y=-x}} \right.\\\\\\\left \{ {{2x=4} \atop {y=-x}} \right. \\\\\\\left \{ {{x=2} \atop {y=-2}} \right. \\\\\\Otvet \ : \ C) \ \ (2 \ ; \ -2)[/tex]