Đáp án:
Tham khảo
Giải thích các bước giải:
2.
$a)\frac{x²+x+1}{x³-1}$
$=\frac{x²+x+1}{(x-1)(x²+x+1)}$
$=\frac{1}{x-1}$
0$b)\frac{2x²-12x+18}{2xy-6y}$
$=\frac{2(x²-6x+9)}{2y(x-3)}$
$=\frac{2(x-3)²}{2y(x-3)}$ $0
$=\frac{2(x-3)}{2y}$
$=\frac{x-3}{y}$
$c)\frac{x²-4y²+2x+1}{4x²+4x-8xy}$
$=\frac{(x²+2x+1)-4y²}{4x(x+1-2y)}$
$=\frac{(x+1)²-(2y)²}{4x(x+1-2y)}$
$=\frac{(x+1-2y)(x+1+2y)}{4x(x+1-2y)}$
$=\frac{x+1+2y}{4x}$
$d)\frac{2x²-7x+6}{x²-2x-3}$
$=\frac{2x²-3x-4x+6}{x²+3x-x-3}$
$=\frac{(2x²-4x)-(3x-6)}{(x²-x)-(3x-3)}$
$=\frac{2x(x-2)-3(x-2)}{x(x-1)-3(x-1)}$
$=\frac{(x-2)(2x-3)}{(x-1)(x-3)}$
