(x+1)/99 + (x+2)/98 + (x+3)/97 + (x+4)/96 = -4
=> [(x+1)/99 +1] +[(x+2)/98+1]+[(x+3)/97+1]+[(x+4)/96+1] = 0
=>[(x+100)/99] + [(x+100)/98] +[(x+100)/97] + [(x+100)/96]=0
=>(x+100)(1/99+1/98+1/97+1/96)=0
=>x+100=0
=>x= -100
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
x-1/99+x-2/98+x-3/97+x-4/96-4=0
x+1/99+x+2/98+x+3/97+x+4/96=-4
Tìm x bt:
a) x-1/99 + x-2/98 + x-3/97 + x-4/96 = 4
b) x+1/99 + x+2/98 + x+3/97 = 3
c) x-1/99 + x-2/49 + x-4/32 = 6
Giúp mik với! Th5 mik mới nộp nhưng mong các bn giúp mik!
Tim x
x+1/99+x+2/98+x+3/97+x+4/96=-4
x+1/99+x+2/98+x+3/97+x+4/96=(-4)
Tìm x biết : (x-1)/ 99 + (x-2) /98 + (x-3) /97 + (x-4) /96 + (x-5) / 95 = 5
Tính:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
x+1/99+x+1/98+x+1/97+x+1/96+x=-4
