Đáp án:
Giải thích các bước giải:
a) $x-2-\frac{x^2-10}{x+2}=\frac{x^2-4-x^2+10}{x+2}=\frac{6}{x+2}$
b) $x^2+y^2-2.\frac{x^4+y^4}{x^2+y^2}=\frac{(x^2+y^2)^2-2(x^4+y^4)}{x^2+y^2}=\frac{-(x^4-2x^2y^2+y^4)}{x^2+y^2}=\frac{-(x^2-y^2)^2}{x^2+y^2}$
c) $\frac{x-3}{4x+4}-\frac{x-1}{6x-30}=\frac{3(x-3)(x-5)-2(x-1)(x+1)}{12(x+1)(x-5)}=\frac{3x^2-24x+45-2x^2+2}{12(x+1)(x-5)}=\frac{x^2-24x+49}{(x+1)(x-5)}$
d) $\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}=\frac{1}{x(1-5x)}+\frac{25x-15}{(1-5x)(1+5x)}=\frac{1+5x+x(25x-15)}{(1-5x)(1+5x)x}=\frac{25x^2-10x+1}{(1-5x)(1+5x)x}=\frac{(5x-1)^2}{(1-5x)(1+5x)x}=\frac{5x-1}{(1+5x)x}$
e) $\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{(x-3y)(x+3y)}-\frac{3y}{x(x+3y)}=\frac{x(x+9y)-3y(x-3y)}{x(x-3y)(x+3y)}=\frac{x^2+6xy+9y^2}{x(x+3y)(x-3y)}=\frac{(x+3y)^2}{x(x+3y)(x-3y)}=\frac{x+3y}{x(x-3y)}$
f) $\frac{1}{1+x}-\frac{1}{x^3+1}+\frac{1}{x^2-x+1}=\frac{1}{1+x}-\frac{1}{(x+1)(x^2-x+1)}+\frac{1}{x^2-x+1}=\frac{x^2-x+1-1+(x+1)}{(x+1)(x^2-x+1)}=\frac{x^2+1}{(x+1)(x^2-x+1)}$
