Tìm x x+2017/2+x+2017/6+x+2017/12=0 (0.1.2.3.4…999+100%).x^2=64

0 bình luận về “Tìm x x+2017/2+x+2017/6+x+2017/12=0 (0.1.2.3.4…999+100%).x^2=64”

1. Đáp án:

   +`x=(-2017)`

   +\(\left[ \begin{array}{l}x=8\\x=-8\end{array} \right.\) 

   Giải thích các bước giải:

   `**[x+2017]/2+[x+2017]/6+[x+2017]/12=0`
   `(x+2017).(1/2+1/6+1/12)=0`
   `(x+2017)(1/1.2+1/2.3+1/3.4)=0`
   `(x+2017)(1-1/2+1/2-1/3+1/3-1/4)=0`
   `(x+2017)(1-1/4)=0`
   `(x+2017)(4/4-1/4)=0`
   `(x+2017).3/4=0`
   Vì `3/4 !=0`
   `=>x+2017=0`
   `x=0-2017`
   `x=(-2017)`
   Vậy `x=(-2017)`

   `**(0.1.2.3.4…999+100%).x^2=64`
   `(0+100/100).x^2=64`
   `1.x^2=64`
   `x^2=64`
   Vì mũ chẵn sẽ có 2 đáp án.
   `=>`\(\left[ \begin{array}{l}x^2=8^2\\x=(-8)^2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=8\\x=-8\end{array} \right.\) 
   Vậy \(\left[ \begin{array}{l}x=8\\x=-8\end{array} \right.\) 

   Bình luận

2. Giải thích các bước giải:

   `a)`
   `(x+2017)/2+(x+2017)/6+(x+2017)/12=0`
   `=>(x+2017) . 1/2+(x+2017). 1/6+(x+2017). 1/12=0`
   `=>(x+2017)(1/2+1/6+1/12)=0`
   Vì `1/2+1/6+1/12=6/12+2/12+1/12=9/12\ne0` nên
   `x+2017=0`
   `=>x=0-2017`
   `=>x=-2017`
   Vậy `x=-2017`
   `b)`
   `(0.1.2.3.4…..999+100%).x^2=64`
   `=>(0+100/100).x^2=64`
   `=>100/100 . x^2=64`
   `=>1.x^2=64`
   `=>x^2=64/1`
   `=>x^2=64`
   `=>x^2=(+-8)^2`
   `=>x=+-8`
   Vậy `x=+-8`

   Bình luận