[Toán 8] Giải phương trình

T

thanhlan9

Đk x# 49,50
Đặt [TEX]a= \frac{x-49}{50}, b= \frac{x-50}{49}[/TEX]
Ta có a+b= [TEX]\frac{1}{a} + \frac{1}{b} \Leftrightarrow a+b=0 , ab=1[/TEX]
Ban tự giải nốt nha
 
T

thaotran19

$\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}$
\Leftrightarrow $(\dfrac{x-49}{50}-1)+(\dfrac{x-50}{49}-1)=(\dfrac{49}{x-50}-1)+(\dfrac{50}{x-49}-1)$
\Leftrightarrow $\dfrac{x-99}{50}+\dfrac{x-99}{49}+\dfrac{x-99}{x-50}+\dfrac{x-99}{x-49}=0$
\Leftrightarrow $(x-99)(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{x-50}+\dfrac{1}{x-49})=0$
\Leftrightarrow $x-99=0$(vì $\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{x-50}+\dfrac{1}{x-49} \not= 0)$
\Leftrightarrow $x=99$
 
N

ninjatapsu

$\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}$
\Leftrightarrow $(\dfrac{x-49}{50}-1)+(\dfrac{x-50}{49}-1)=(\dfrac{49}{x-50}-1)+(\dfrac{50}{x-49}-1)$
\Leftrightarrow $\dfrac{x-99}{50}+\dfrac{x-99}{49}+\dfrac{x-99}{x-50}+\dfrac{x-99}{x-49}=0$
\Leftrightarrow $(x-99)(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{x-50}+\dfrac{1}{x-49})=0$
\Leftrightarrow $x-99=0$(vì $\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{x-50}+\dfrac{1}{x-49} \not= 0)$
\Leftrightarrow $x=99$
Hình như bạn nhầm rồi.
$\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{x-50}+\dfrac{1}{x-49} =0$ \Leftrightarrow x=0(TMDK) nữa mà
 
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