=>\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
=>\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
=>\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
=>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
Cả lời giải nhé!
Tìm x biết : (x-1)/ 99 + (x-2) /98 + (x-3) /97 + (x-4) /96 + (x-5) / 95 = 5
Tính:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
Tìm x, biết:
a) 1 1/5 . x + 2/3 . x = - 56/125
b) x+1/99 + x+2/98 + x+3/97 + x+4/96 = - 4
