\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
Giải pt sau:
1,x+2/2002 +x+5/1999 +x+201/1803=-3
2,x+1/99 +x+3/97 +x+5/95=x+9/91 +x+8/92 +x+7/93.
[(x+1)/99]+[(x+3)/97]+[(x+5)/95]= [(x+7)/93]+[(x+9)/91]+[(x+11)/89]
giải PT: a, (4x-5)2 (2x-3)(x-1)=9
b,\(\frac{5}{x-8}+1=\frac{23}{x^2-5x-24}+\frac{2}{x+3}\)
c,(\(\left(\frac{x-1}{99}+\frac{x-99}{1}\right)+\left(\frac{x-3}{97}+\frac{x+97}{3}\right)+\left(\frac{x-5}{93}+\frac{x-95}{5}\right)=6\)
Giải phương trình sau:
a) x+1/2004 + x+2/2003 = x+3/2002 + x+4/2001
b) 201-x/99 + 203-x/97 + 205-x/95 + 3 = 0
(x-1/99+x-99)+(x-3/97+x-7/93)+(x-5/95+x-95/5)=6
a,(x-1/99+x-99)+(x-3/97+x-7/93)+(x-5/95+x-95/5)=6
\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
Giải phương trình sau
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
