Toán Lớp 8: Giải Phg Trình
b, $\frac{x+6}{1999}$+ $\frac{x+8}{1997}$ =$\frac{x+10}{1995}$ +$\frac{x+12}{1993}$
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Toán Lớp 8: Giải Phg Trình
b, $\frac{x+6}{1999}$+ $\frac{x+8}{1997}$ =$\frac{x+10}{1995}$ +$\frac{x+12}{1993}$
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Giải đáp:
b, (x + 6)/1999 + (x + 8)/1997 = (x + 10)/1995 + (x + 12)/1993
↔ (x + 6)/1999 + 1+ (x + 8)/1997 +1 = (x + 10)/1995 +1 + (x + 12)/1993 + 1
↔ (x + 6 + 1999)/1999 + (x + 8+ 1997)/1997 = (x + 10 + 1995)/1995 + (x + 12+ 1993)/1993
↔ (x + 2005)/1999 + (x + 2005)/1997 = (x + 2005)/1995 + (x + 2005)/1993
↔ (x + 2005)/1999 + (x + 2005)/1997 – (x + 2005)/1995 – (x + 2005)/1993 = 0
↔ (x + 2005).(1/1999 + 1/1997 – 1/1995 – 1/1993) = 0
↔ x + 2005 = 0 (1/1999 + 1/1997 – 1/1995 – 1/1993 \ne 0)
↔ x = -2005
Vậy S = {-2005}
$#dariana$
(x+6)/1999 + (x+8)/1997 =(x+10)/1995 + (x+12)/1993
<=>(x+6+1999)/1999 + (x+8+1997)/1997 = (x+10+1995)/1995 + (x+12+1993)/1993
<=>(x+2005)/1999 + (x+2005)/1997 = (x+2005)/1995 + (x+2005)/1993
<=> (x+2005)/1999 + (x+2005)/1997 – (x+2005)/1995 – (x+2005)/1993=0
<=>(x+2005)(1/1999 +1/1997 – 1/1995 – 1/1993)=0
Do 1/1999 + 1/1997 – 1/1995 – 1/1993 \ne 0
<=>x+2005=0
<=>x=-2005
Vậy S={-2005}