Question

73 даю!!!!
1142(б,в)
1143(б)очееень надо!!!!

Answer 1

$1142)\; ...=\frac{\frac{x+y}{x-y}+\frac{x-y}{x+y}-2}{\frac{x+y}{x-y}-\frac{x-y}{x+y}}=\frac{(x+y)^2+(x-y)^2-2(x-y)(x+y)}{(x+y)^2-(x-y)^2}=\\\\=\frac{x^2+2xy+y^2+x^2-2xy+y^2-2x^2+2y^2}{x^2+2xy+y^2-x^2+2xy-y^2}=\frac{4y^2}{4xy}=\frac{y}{x}$

$1142)\; \; ...=\frac{1+\frac{1}{a+b}}{1-\frac{1}{a+b}}\cdot \frac{(a+b-1)^2}{(a+b+1)^2}=\frac{a+b+1}{a+b-1}\cdot \frac{(a+b-1)^2}{(a+b+1)^2}=\frac{a+b-1}{a+b+1}\\\\\\1143)\; \; ...=\frac{\frac{3}{c^2+cx+x^2}-\frac{1}{c^2-cx+x^2}}{(c-x)^2(c^4+c^2x^2+x^4)}=\\\\=\frac{3c^2-3cx+3x^2-c^2+cx+x^2}{(c-x)(c^2+cx+x^2)(c-x)(c^2-cx+x^2)(c^2+c^2x^2+x^4)}=\\\\=\frac{2(c^2-cx+x^2)(c+x)}{(c^3-x^3)(c-x)(c+x)(c^2-cx+x^2)(c^4+c^2x^2+x^4)}=\\\\=\frac{2(c+x)}{(c^3-x^3)(c^2-x^2)(c^4+c^2x^2+x^4)}=$

$=\frac{2(c+x)}{(c^3-x^3)(c^6-x^6)}=\frac{2(c+x)}{(c^3-x^3)((c^3)^2-(x^3)^2)}=\\\\=\frac{2(c+x)}{(c^3-x^3)(c^3-x^3)(c^3+x^3)}=\frac{2(c+x)}{(c^3-x^3)^2(c^3+x^3)}$