a, .... (đề bài)
<=> $\frac{x-15}{100}$+ $\frac{x-10}{105}$+ $\frac{x-105}{10}$+$\frac{x-100}{15}$ -4=0
<=> ($\frac{x-15}{100}$-1)+($\frac{x-10}{105}$-1)+($\frac{x-105}{10}$-1)+($\frac{x-100}{15}$-1)=0
<=> $\frac{x-115}{100}$+ $\frac{x-115}{105}$+ $\frac{x-115}{10}$+$\frac{x-115}{15}$=0
<=> (x-115)($\frac{1}{100}$+ $\frac{1}{105}$+ $\frac{1}{10}$+$\frac{1}{15}$)=0
<=> x- 115 = 0 => x = 115
vậy.......
b, ...... (đề bài)
<=> ($\frac{x+2000}{20}$+1)+($\frac{x+2010}{10}$)=( $\frac{x-120}{2140}$+1)+($\frac{x-100}{2120}$+1)
<=> $\frac{x+2020}{20}$+$\frac{x+2020}{10}$= $\frac{x+2020}{2140}$+$\frac{x+2020}{2120}$
<=> (x+2020)($\frac{1}{20}$+$\frac{1}{10}$- $\frac{1}{2140}$ - $\frac{1}{2120}$) = 0
<=> x + 2020 = 0 <=> x = -2020
vậy......
