tìm x  a)1/1*2+1/2*3+...+1/x*(x+1)=2021/2022 b) x+1/99+x+2/98+x+3/97+x+4/86=-4 ai trả lời giúp mik với mik đang gấp  
các Bạn ơi, mình đang bí bài này quá, ai giỏi giúp mình với! Cảm ơn cả nhà

a: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)

=>\(\dfrac{1}{x+1}=1-\dfrac{2021}{2022}=\dfrac{1}{2022}\)

=>x+1=2022

=>x=2021

b: Sửa đề: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

=>\(\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\)

=>\(\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

=>\(\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

=>x+100=0

=>x=-100