Question

Bài 1: Tính giá trị các biểu thức sau

a) \( A = \frac{8}{(x^2+3)(x^2-1)} + \frac{2}{x^2+3} + \frac{1}{x-1} \)

b) \( B = \frac{x^3+x^2-2x-20}{x^2-4} - \frac{5}{x+2} + \frac{3}{x-2} \)

c) \( N = \left( \frac{x-y}{x+y} + \frac{x+y}{x-y} \right) \cdot \left( \frac{x^2+y^2}{2xy} + 1 \right) \cdot \frac{xy}{x^2+y^2} \)

Answer 1

a, `A=\frac{8}{(x^2+3)(x^2-1)}+\frac{2}{x^2+3}+\frac{1}{x-1}`

`=\frac{8}{(x^2+3)(x-1)(x+1)}+\frac{2(x-1)(x+1)}{(x^2+3)(x-1)(x+1)}+\frac{(x^2+3)(x+1)}{(x^2+3)(x-1)(x+1)}`

`=\frac{8+2x^2-2+x^3+x^2+3x+3}{(x^2+3)(x-1)(x+1)}`

`=\frac{x^3+3x^2+3x+9}{(x^2+3)(x-1)(x+1)}`

`=\frac{(x^2+3)(x+3)}{(x^2+3)(x-1)(x+1)}`

`=\frac{x+3}{(x-1)(x+1)}`

b, `B=\frac{x^3+x^2-2x-20}{x^2-4}-\frac{5}{x+2}+\frac{3}{x-2}`

`=\frac{x^3+x^2-2x-20}{(x-2)(x+2)}-\frac{5(x-2)}{(x-2)(x+2)}+\frac{3(x+2)}{(x-2)(x+2)}`

`=\frac{x^3+x^2-2x-20-5x+10+3x+6}{(x-2)(x+2)}`

`=\frac{x^3+x^2-4x-4}{(x-2)(x+2)}`

`=\frac{(x+1)(x-2)(x+2)}{(x-2)(x+2)}`

`=x+1`

c, `N=(\frac{x-y}{x+y}+\frac{x+y}{x-y})*(\frac{x^2+y^2}{2xy}+1)*\frac{xy}{x^2+y^2}`

`=\frac{(x-y)^2+(x+y)^2}{(x+y)(x-y)}*\frac{x^2+y^2+2xy}{2xy}*\frac{xy}{x^2+y^2}`

`=\frac{2(x^2+y^2)}{(x+y)(x-y)}*\frac{(x+y)^2}{2xy}*\frac{xy}{x^2+y^2}`

`=\frac{x+y}{x-y}`