3/(x^2+5x+6) + 3/(x^2+7x+12) + 3/(x^2+9x+20) + 3/(x^2+11x+30)=1/5

2 bình luận về “3/(x^2+5x+6) + 3/(x^2+7x+12) + 3/(x^2+9x+20) + 3/(x^2+11x+30)=1/5”

1. Lời giải và giải thích chi tiết:

   3/(x^2+5x+6)+3/(x^2+7x+12)+3/(x^2+9x+20)+3/(x^2+11x+30)=1/5(x\ne-2;-3;-4;-5;-6)
   <=>3/((x+2)(x+3))+3/((x+3)(x+4))+3/((x+4)(x+5))+3/((x+5)(x+6))=1/5
   <=>3.(1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)+1/(x+5)-1/(x+6))=1/5
   <=>3.(1/(x+2)-1/(x+6))=1/5
   <=>1/(x+2)-1/(x+6)=1/15
   <=>(x+6-x-2)/((x+2)(x+6))=1/15
   <=>(4/((x+2)(x+6))=4/(60)
   =>(x+2)(x+6)-60=0
   <=>x^2+8x+12-60=0
   <=>x^2+8x-48=0
   <=>[(x=4),(x=-12):}
   Vây S={4;-12}

   Trả lời
2. 3/(x^2 + 5x+6) + 3/(x^2 + 7x+12) + 3/(x^2+9x+20) + 3/(x^2+11x+30) = 1/5

   <=> 3/[(x+2)(x+3)] + 3/[(x+3)(x+4)] + 3/[(x+4)(x+5)] + 3/[(x+5)(x+6)] = 1/5 ĐKXĐ:x≠-2;-3;-4;-5;-6

   <=> 3/(x+2) – 3/(x+3) + 3/(x+3) – 3/(x+4) + 3/(x+4) – 3/(x+5) + 3/(x+5) – 3/(x+6) = 1/5

   <=> 3/(x+2) – 3/(x+6) = 1/5 MTC: 5(x+2)(x+6)

   <=> 15(x+6) – 15(x+2) = (x+2)(x+6)

   <=> 15x + 90 – 15x – 30  =x^2 + 8x + 12

   <=> 0 = x^2 + 8x – 48

   <=> x^2 – 4x + 12x – 48 =0

   <=> x(x-4) + 12(x-4) =0

   <=>(x-4)(x+12)=0

   <=>\(\left[ \begin{array}{l}x-4=0\\x+12=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=4\\x=-12\end{array} \right.\) ™

   Vậy S={4;-12}

   $\color{lightblue}{eriet}$

   Trả lời