Question

[Phạm Tuấn Long]

I : Giải các phương trình sau :

a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)

b) \(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

c) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

d) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

e) \(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)

f) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

g) \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

các bạn ơi giúp mình với

mình cảm ơn

Answer 1

[Nguyễn Thành Trương]

Câu a)

Answer 2

[Nguyễn Thành Trương]

b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100

Vậy x = 100