1. $A=$$\frac{4x^{2}+8x+4}{x+1}$
a) Biểu thức A có nghĩa khi $x+1\neq0$ →$x\neq-1$
b)$A=$$\frac{4x^{2}+8x+4}{x+1}$
$A=$$\frac{4(x^{2}+2x+1)}{x+1}$
$A=$$\frac{4(x^{2}+2.x.1+1^{2})}{x+1}$
$A=$$\frac{4(x+1)^{2}}{x+1}$
$A=4(x+1)$
Thế $x=\frac{1}{4}$ vào $A$
$A=4.(\frac{1}{4}+1)$
$A=4.\frac{5}{4}$
$A=5$
2. $B=(\frac{1}{x+1}+\frac{1}{x-1}):\frac{2x}{1-x^{2}}$
$B=(\frac{1.(x-1)}{(x+1)(x-1)}+\frac{1.(x+1)}{(x-1)(x+1)}).\frac{1-x^{2}}{2x}$
$B=\frac{x-1+x+1}{x^{2}-1}.\frac{x^{2}-1}{-2x}$
$B=\frac{2x}{x^{2}-1}.\frac{x^{2}-1}{-2x}$
$B=-1$
