Giải thích các bước giải:
a.Ta có:
$\dfrac{1}{1.2}+\dfrac1{2.3}+...+\dfrac{1}{x(x+1)}=\dfrac{28}{29}$
$\to \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{(x+1)-x}{x(x+1)}=\dfrac{28}{29}$
$\to \dfrac11-\dfrac12+\dfrac12-\dfrac13+...+\dfrac1x-\dfrac1{x+1}=\dfrac{28}{29}$
$\to 1-\dfrac1{x+1}=\dfrac{28}{29}$
$\to\dfrac1{x+1}=\dfrac1{29}$
$\to x+1=29$
$\to x=28$
b.Ta có:
$\dfrac{1}{12}+\dfrac1{20}+\dfrac{1}{30}+\dfrac1{42}+...+\dfrac1{x(x+1)}=\dfrac{34}{35}$
$\to \dfrac{1}{3.4}+\dfrac1{4.5}+\dfrac{1}{5.6}+\dfrac1{6.7}+...+\dfrac1{x(x+1)}=\dfrac{34}{35}$
$\to \dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{(x+1)-x}{x(x+1)}=\dfrac{34}{35}$
$\to \dfrac13-\dfrac14+\dfrac14-\dfrac15+\dfrac15-\dfrac16+\dfrac16-\dfrac17+...+\dfrac1x-\dfrac1{x+1}=\dfrac{34}{35}$
$\to \dfrac13-\dfrac1{x+1}=\dfrac{34}{35}$
$\to \dfrac1{x+1}=-\dfrac{67}{105}$
$\to$Không tồn tại $x$ thỏa mãn đề
