$\text{(a + b + c – d)(a – b – c – d) = ( a + b – c + d)(a – b + c + d)}$
$\text{⇒ $\dfrac{a + b + c – d}{a + b – c + d}$ = $\dfrac{a – b + c + d}{a – b – c – d}$ ⇔ $\dfrac{(a+b) + (c-d)}{(a-b) – (c+d)}$ = $\dfrac{(a+b) + (c+d)}{(a-b) – (c+d)}$}$
$\text{Đặt A = a + b ; B = c – d ; C = a – b ; D = c + d. Ta được:}$
Đáp án:
$\text{(a + b + c – d)(a – b – c – d) = ( a + b – c + d)(a – b + c + d)}$
$\text{⇒ $\dfrac{a+b}{a-b}$ = $\dfrac{c-d}{c+d}$}$
Giải thích các bước giải:
$\text{Giải:}$
$\text{Ta có : }$
$\text{(a + b + c – d)(a – b – c – d) = ( a + b – c + d)(a – b + c + d)}$
$\text{⇒ $\dfrac{a + b + c – d}{a + b – c + d}$ = $\dfrac{a – b + c + d}{a – b – c – d}$ ⇔ $\dfrac{(a+b) + (c-d)}{(a-b) – (c+d)}$ = $\dfrac{(a+b) + (c+d)}{(a-b) – (c+d)}$}$
$\text{Đặt A = a + b ; B = c – d ; C = a – b ; D = c + d. Ta được:}$
$\text{$\dfrac{A + B}{A – B}$ = $\dfrac{C + D }{C – D}$ ⇒ $\dfrac{A}{B}$ = $\dfrac{C}{D}$ ⇔ $\dfrac{a + b}{c-d}$ = $\dfrac{a-b}{c+d}$ }$
$\text{⇒ $\dfrac{a+b}{a-b}$ = $\dfrac{c-d}{c+d}$}$
$\text{Ta được :}$
$\text{(a + b + c – d)(a – b – c – d) = ( a + b – c + d)(a – b + c + d)}$
$\text{⇒ $\dfrac{a+b}{a-b}$ = $\dfrac{c-d}{c+d}$}$