tìm x dạng khó. Ai thấy mình giỏi thì vào!

H

hoailinhminho

$\dfrac{x-49}{50} + \dfrac{x-50}{49} =\dfrac{50}{x-49} + \dfrac{49}{x-50} \\ \Longrightarrow \dfrac{49(x-49)}{49.50} + \dfrac{50(x-50)}{49.50} =\dfrac{50(x-50)}{(x-49)(x-50)} + \dfrac{49(x-49)}{(x-50)(x-49)} \\ \Longrightarrow \dfrac{49x - 49^2 + 50x - 50^2}{50 . 49} =\dfrac{50x - 50^2 + 49x - 49^2}{x^2 - 49x - 50x + 49.50} \\ \Longrightarrow \dfrac{49(x-49)}{49.50} + \dfrac{50(x-50)}{49.50} =\dfrac{50(x-50)}{x-49(x-50)} + \dfrac{49(x-49)}{x-50(x-49)} \\ \Longrightarrow\dfrac{49x - 49^2 + 50x - 50^2}{50 . 49}=\dfrac{50x - 50^2 + 49x - 49^2}{x^2 - 49x - 50x + 49.50} $

vì $49x - 49^2 + 50x - 50^2 = 50x - 50^2 + 49x - 49^2 \\ \Longrightarrow 50.49 = x^2 - 49x - 50x + 49.50 \\ \Longrightarrow 50.49-49.50= x^2 - 99x = x(x-99)=0$

$\Longrightarrow \left[ \begin{array}{ll} x=0 \\ x-99=0 \Longrightarrow x=99 \end{array} \right.$

Chú ý đánh latex. Xem thêm tại đây.
Ps: Đã sửa!
 
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L

lovelybones311

$\dfrac{x-49}{50} + \dfrac{x-50}{49} =\dfrac{50}{x-49} + \dfrac{49}{x-50} $

Đặt
$ a = x-49$
$ b = x-50$

=> $\dfrac{a}{50} + \dfrac{b}{49} = \dfrac{50}{a} +\dfrac{49}{b} $

=> $\dfrac{49a+50b}{50.49}=\dfrac{49a+50b}{ab}$

=> $ ab = 50.49 $

Mặt khác : $a-b = x-49 - (x-50) =1 $
=> $a=b+1$

Nên : $b.(b+1) -50.49 =0$

=> $b^2 + b -50.49 =0$
=> $b=49 hay b=-50$
=> $x=0 hay x=99 $
 
N

nguyehuuhuy14112000

$\dfrac{x-49}{50} + \dfrac{x-50}{49} =\dfrac{50}{x-49} + \dfrac{49}{x-50} $

=> $\dfrac{(x-49).49+50.(x-50)}{49.50} = \dfrac{50.(x-50)+(x-49).49}{(x-49).(x-50)} $

=> $ (x-49).(x-50)= 50.49 $

$ = x-49 - (x-50) =1 $

$=>x-49=(x-50)+1$

Nên : ${(x-50).[(x-50)+1)} -50.49 =0$

$TH1:=>x-50=49 $

$=>x=99$

$TH2: x-50=-50$

$=>x=0$
 
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