Cho A=(x/x+5+5/x-5+10x/x^2-25) (1-5/x)

a) rút gọn A

b) tính A khi x=10

$ĐKXĐ:x\neq 0;-5;-5$

$a,$ 

$A=(\dfrac{x}{x+5}+\dfrac{5}{x-5}+\dfrac{10x}{x^2-25})(1-\dfrac{5}{x})$ 

$A=(\dfrac{x}{x+5}+\dfrac{5}{x-5}+\dfrac{10x}{(x-5)(x+5)})(\dfrac{x-5}{x})$ 

$A=(\dfrac{x(x-5)}{(x-5)(x+5)}+\dfrac{5(x+5)}{(x+5)(x-5)}+\dfrac{10x}{(x-5)(x+5)})(\dfrac{x-5}{x})$ 

$A=(\dfrac{x^2-5x}{(x-5)(x+5)}+\dfrac{5x+25}{(x+5)(x-5)}+\dfrac{10x}{(x-5)(x+5)})(\dfrac{x-5}{x})$ 

$A=(\dfrac{x^2+10x+25}{(x-5)(x+5)})(\dfrac{x-5}{x})$ 

$A=(\dfrac{(x+5)^2}{(x-5)(x+5)})(\dfrac{x-5}{x})$ 

$A=\dfrac{x+5}{x-5}.\dfrac{x-5}{x}$ 

$A=\dfrac{x+5}{x}$ 

$b,$ Khi $x=10$

$A=\dfrac{10+5}{10}=\dfrac{15}{10}=\dfrac{3}{2}$

Vậy $A=\dfrac{3}{2}↔x=10$

Đáp án`+`Giải thích các bước giải:

`a)`

`A = (x/{x + 5} + 5/{x - 5} + {10x}/{x^2 - 25}) (1 - 5/x)` (ĐKXĐ: `x ne +- 5 ; x ne 0`)

`= (x/{x + 5} + 5/{x - 5} + {10x}/{(x - 5)(x + 5)}) (x/x - 5/x)`

`= ({x(x - 5)}/{(x - 5)(x + 5)} + {5(x + 5)}/{(x - 5)(x + 5)} + {10x}/{(x - 5)(x + 5)}) . {x - 5}/x`

`= ({x^2 - 5x}/{(x - 5)(x + 5)} + {5x + 25}/{(x - 5)(x + 5)} + {10x}/{(x - 5)(x + 5)}) . {x - 5}/x`

`= {x^2 - 5x + 5x + 25 + 10x}/{(x - 5)(x + 5)} . {x - 5}/x`

`= {x^2 + 25 + 10x}/{1 . (x + 5)} . 1/x`

`= {x^2 + 10x + 25}/{x + 5} . 1/x`

`= {(x + 5)^2}/{x + 5} . 1/x`

`= {x + 5}/1 . 1/x`

`= {x + 5}/x`

`b)`

Thay `x = 10` vào biểu thức `A` ta có:

`{10 + 5}/10 = 15/10 = 3/2`

Vậy `A = 3/2` khi `x = 10`