tìm x a) (x+1) + (x+3)+…+(x+99)=0 b)(x^2 +1) + (x-10) <0 c) (x^2 + 2)(x+5) >0 giúp mik với

` a) ` ` (x + 1) + (x + 3) + (x + 5) + … + (x + 99) = 0 `

` => x + 1 + x + 3 + x + 5 + … + x + 99 = 0 `

` => (x + x + x + … + x) + (1 + 3 + 5 + … + 99) = 0 `

` => 50x + [(1 + 99) + (3 + 97) + … + (49 + 51)] = 0 `

` => 50x + [100 . 25] = 0 `

` => 50x + 2500 = 0 `

` => 50x = -2500 `

` => x = -50 `

` b) ` ` (x^{2} + 1)(x – 10) < 0 `

` => ` \(\left[ \begin{array}{l}\left \{ {{x^{2}+1>0} \atop {x-10<0}} \right. \\\left \{ {{x^{2}+1<0} \atop {x-10>0}} \right. \end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}\left \{ {{x^{2}>-1(loại)} \atop {x<10(chọn)}} \right. \\\left \{ {{x^{2}<-1(loại)} \atop {x>10(loại)}} \right. \end{array} \right.\) 

` => x < 10 `

Vậy ` x < 10 `

` c) ` ` (x^{2} + 2)(x + 5) > 0 `

` => ` \(\left[ \begin{array}{l}\left \{ {{x^{2}+2>0} \atop {x+5>0}} \right. \\\left \{ {{x^{2}+2<0} \atop {x+5<0}} \right. \end{array} \right.\) 

` => ` \(\left[ \begin{array}{l}\left \{ {{x^{2}>-2(loại)} \atop {x>-5(loại)}} \right. \\\left \{ {{x^{2}<-2(loại)} \atop {x<-5(chọn)}} \right. \end{array} \right.\) 

` => x < -5 `