Question

ПОЖАЛУЙСТА ПОМОГИТЕ!!!

Answer 1

$((\frac{5a-4}{a-4}-\frac{a-5}{a-1})*\frac{a-4}{a-2}+\frac{4a(a-1)}{a&2-2a+1} ) *\frac{a-1}{a+1}$

$1)\frac{5a-4}{a-4}-\frac{a-5}{a-1}=\frac{(5a-4)(a-1)}{(a-4)(a-1)}-\frac{(a-5)(a-4)}{(a-1)(a-4)}=\frac{(5a-4)(a-1)-(a-5)(a-4)}{(a-4)(a-1)}=\\=\frac{(5a*a+5a*(-1)-4*a-4*(-1))-(a*a+a*(-4)-5*a-5*(-4)}{(a-4)(a-1)}=\frac{5a^2-5a-4a+4-(a^2-4a-5a+20)}{(a-4)(a-1)} =\frac{5a^2-9a+4-a^2+9a-20}{(a-4)(a-1)} =\frac{4a^2-16}{(a-4)(a-1)} =\frac{4(a^2-4)}{(a-4)(a-1)}=\frac{4(a-2)(a+2)}{(a-4)(a-1)}$

$2)\frac{4(a-2)(a+2)}{(a-4)(a-1)}*\frac{a-4}{a-2} =\frac{4(a-2)(a+2)(a-4)}{(a-4)(a-1)(a-2)}=\frac{4(a+2)}{(a-1)}$

$3)\frac{4(a+2)}{(a-1)}+\frac{4a(a-1)}{a^2-2a+1} =\frac{4(a+2)}{(a-1)}+\frac{4a(a-1)}{(a-1)^2}=\frac{4(a+2)(a-1)}{(a-1)^2}+\frac{4a(a-1)}{(a-1)^2}=\frac{4(a+2)(a-1)+4a(a-1)}{(a-1)^2}=\frac{4(a^2+a*(-1)+2a+2*(-1))+4a*a+4a*(-1)}{(a-1)^2}=\frac{4a^2+4a-8+4a^2-4a}{(a-1)^2} =\frac{8a^2-8}{(a-1)^2}=\frac{8(a^2-1)}{(a-1)^2}=\frac{8(a-1)(a+1)}{(a-1)^2}=\frac{8(a+1)}{(a-1)}$

$4)\frac{8(a+1)}{(a-1)}*\frac{a-1}{a+1} =\frac{8(a+1)(a-1)}{(a-1)(a+1)}=8$