I)(4y²-3y)/(2y²+4)=y(4y-3)/(y+2)(y+2)
при y=-2: -2(4·(-2)-3)/(-2+2)(-2+2); -2(-8-3)/0·0; -2·(-11)/0; 22/0
Ответ: при y=(0;-2)
II)a)(7c-21d)²/(2c-6d)=(7c-21d)(7c-21d)/2(c-3d)=(49c²-147dc-147dc+441d²)/2(c-3d)= (49c²-294dc+441d²)/2(c-3d)=7(c-3d)(c+3d)/2(c-3d)=7(c+3d)/2=(7c+21d)/2
b)(y³·y³·y²-2·y²·y³·y²+y²·y²·y²)/(7·y²·y³·y³·y-7·y²·y³·y²)=y²(y³·y³-2·y³·y²+y²·y²)/
/(y²(7·y³·y³·y-7·y³·y²)=(y³·y³-2·y³·y²+y²·y²)/(7·y³·y³·y-7·y·y²·y²)=(y³-y²)(y³-y²)/(7y(y³-y² )(y³+y²)=(y³-y²)/(7y(y³+y²)=(y³-y²)/(7y·y³+7y³)
III)a)(x+12)/(16-x²)+4/(4x+x²)=(x+12)/(4-x)(4+x)+4/(x(4+x))=x(x+12)/x(4-x)(4+x)+4(4-x)/ (4+x)=(x(x+12)+4(4-x))/x(4-x)(4+x)=(x²+12x+16-4x)/x(16-4x+4x-x²)=(x²+8x+16)/(x²+16)=(x+4)(x+4)/(x+4)(x-4)=(x+4)/(x-4)
b)1/(a²b+ab²)+3/(a³+b³)=1/(ab(a+b))+3/(a²-ab+b²)(a+b)=(a²-ab+b²)/(ab(a+b))(a²-ab+b²)+3(ab)/(ab(a+b)(a²-ab+b²)=(a²-ab+b²+3ab)/ab(a³+b³)=(a²+2ab+b²)/(a·a³b+ab·b³)=(a+b)²/(aba³+abb³)