Bài tập toán 6 : Luyện tập phân số và liên quan

a) $\frac{-8}{18}+\frac{-15}{27}=\frac{-4}{9}+\frac{-5}{9}=-1$

b) $\frac{16}{20}-\frac{15}{25}=\frac{4}{5}-\frac{3}{5}=\frac{1}{5}$

a) $-5+\frac{-2}{7}=\frac{-35}{7}+\frac{-2}{7}=\frac{-37}{7}$

b) $\frac{27}{35}-1=\frac{27}{35}-\frac{35}{35}=\frac{-8}{35}$

a) $\frac{3}{8}+\frac{-5}{8}-\frac{1}{8}=\frac{3-5-1}{8}=\frac{-3}{8}$

b) $\frac{3}{4}-\frac{-1}{2}+\frac{1}{6}=\frac{9}{12}+\frac{6}{12}+\frac{2}{12}=\frac{17}{12}$

a) $\frac{-4}{10}+\frac{15}{25}+\frac{-2}{15}=\frac{-2}{5}+\frac{3}{5}+\frac{-2}{15}=\frac{-6}{15}+\frac{9}{15}+\frac{-2}{15}=\frac{1}{15}$

b) $\frac{-7}{12}+\frac{-4}{30}-\frac{-1}{6}=\frac{-35}{60}+\frac{-8}{60}+\frac{10}{60}=\frac{-33}{60}=\frac{-11}{20}$

$A=\frac{5}{11}+\frac{-3}{7}+\frac{6}{11}+\frac{2}{3}+\frac{-4}{7}=\left(\frac{5}{11}+\frac{6}{11}\right)+\left(\frac{-3}{7}+\frac{-4}{7}\right)+\frac{2}{3}=\frac{2}{3}$

$\begin{array}{l}B=\frac{-4}{13}+\frac{3}{5}+\frac{1}{3}-\frac{9}{13}+\frac{2}{5}=\left(\frac{-4}{13}+\frac{-9}{13}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)+\frac{1}{3}\\ =(-1)+1+\frac{1}{3}=0+\frac{1}{3}=\frac{1}{3}\end{array}$

a) 

$\begin{array}{l}x-\frac{1}{2}=\frac{2}{9}+\frac{-1}{5}\\ x-\frac{1}{2}=\frac{10}{45}+\frac{-9}{45}\\ x-\frac{1}{2}=\frac{1}{45}\\ x=\frac{10}{45}+\frac{1}{2}\\ x=\frac{20}{90}+\frac{45}{90}=\frac{65}{90}=\frac{13}{18}\end{array}$

b)

$\begin{array}{l}\frac{x}{10}=\frac{3}{15}-\frac{1}{2}=\frac{-9}{30}=\frac{-3}{10}\\ x=-3\end{array}$

a) 

$\begin{array}{l}x+\frac{3}{7}=\frac{-2}{7}\\ x=\frac{-2}{7}-\frac{3}{7}=\frac{-5}{7}\end{array}$

b)

$\begin{array}{l}\frac{1}{12}+x=\frac{-2}{3}\\ x=\frac{-2}{3}-\frac{1}{12}=\frac{-8}{12}-\frac{1}{12}=\frac{-9}{12}=\frac{-3}{4}\end{array}$

a) 

$\begin{array}{l}\frac{1}{4}-x=\frac{1}{20}+(-2)\\ \frac{1}{4}-x=\frac{-39}{20}\\ x=\frac{1}{4}-\frac{-39}{20}=\frac{5}{20}+\frac{39}{20}=\frac{44}{20}=\frac{11}{5}\end{array}$

b) 

$\begin{array}{l}\frac{-2}{5}-x=\frac{4}{15}+\frac{-1}{3}\\ \frac{-2}{5}-x=\frac{-1}{15}\\ x=\frac{-2}{5}-\frac{-1}{15}=\frac{-6}{15}-\frac{-1}{15}=\frac{-5}{15}=\frac{-1}{3}\end{array}$

$\begin{array}{l}\frac{-6}{5}-\frac{2}{3}-\frac{-7}{15}\le x\le \frac{1}{10}+\frac{4}{5}-\frac{-7}{10}\\ \frac{-18}{15}+\frac{-10}{15}+\frac{7}{15}\le x\le \frac{1}{10}+\frac{8}{10}+\frac{7}{10}\\ \frac{-7}{5}\le x\le \frac{16}{10}\\ \Rightarrow x=-1;0;1\end{array}$

$\begin{array}{l}\frac{-8}{5}+\frac{-3}{5}<\mathrm{...}<\frac{-3}{12}+\frac{1}{4}\\ \Rightarrow \frac{-11}{5}<\mathrm{...}<0\\ \Rightarrow \frac{-11}{5}<-1<0\end{array}$

$\begin{array}{l}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}=1-\frac{1}{7}<1\end{array}$

So sánh: S < 1.

Với mọi số tự nhiên n thì $\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}$

$\begin{array}{l}\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\mathrm{...}+\frac{1}{19.20}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\mathrm{...}+\frac{1}{19}-\frac{1}{20}\\ \frac{1}{2}-\frac{1}{20}=\frac{9}{20}\end{array}$

$\begin{array}{l}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\mathrm{...}+\frac{2}{99.101}\\ =1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\mathrm{...}+\frac{1}{99}-\frac{1}{101}\\ =1-\frac{1}{101}=\frac{100}{101}\end{array}$

$\begin{array}{l}B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\mathrm{...}+\frac{3}{97.100}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}\mathrm{...}+\frac{1}{97}-\frac{1}{100}\\ =1-\frac{1}{100}=\frac{99}{100}\end{array}$

Mỗi giờ vòi  A chảy được $\frac{1}{5}$ bể , vòi B chảy được $\frac{1}{6}$ bể.

Vậy mỗi giờ vòi  A chảy nhiều hơn vòi B là $\frac{1}{5}-\frac{1}{6}=\frac{1}{30}$ bể.

a) Chiều dài đám đất là: $\frac{1}{20}+\frac{1}{4}=\frac{3}{10}\left(km\right)$

b) Nửa chu vi đám đất là: $\frac{1}{20}+\frac{3}{10}=\frac{7}{20}\left(km\right)$

a) $\frac{7}{-78}+\frac{9}{65}=\frac{-35}{390}+\frac{54}{390}=\frac{19}{390}$

b) $-5+\frac{19}{7}=\frac{-35}{7}+\frac{19}{7}=\frac{-16}{7}$

a) $\frac{-7}{8}+\frac{5}{-12}+\frac{3}{4}=\frac{-21}{24}+\frac{-10}{24}+\frac{18}{24}=\frac{-13}{24}$

b) $\frac{21}{49}+\frac{-7}{5}+\frac{34}{85}=\frac{3}{7}+\frac{-7}{5}+\frac{2}{5}=-\frac{4}{7}$

a) $\frac{3}{4}-\left|\frac{-1}{2}\right|=\frac{3}{4}-\frac{1}{2}=\frac{3}{4}-\frac{2}{4}=\frac{1}{4}$

b) $\frac{3}{8}-\frac{4}{5}-\frac{-17}{40}=\frac{15}{40}-\frac{32}{40}+\frac{17}{40}=0$

a) $\frac{-1}{2}-\frac{7}{8}+\frac{13}{16}=\frac{-8}{16}-\frac{14}{16}+\frac{13}{16}=\frac{-9}{16}$

b) $\frac{13}{7}-\frac{-3}{11}+\frac{6}{42}=\left(\frac{13}{7}+\frac{1}{7}\right)+\frac{3}{11}=2+\frac{3}{11}=2\frac{3}{11}$

a) $\frac{5}{8}+\frac{3}{17}+\frac{4}{18}+\frac{20}{-17}+\frac{-2}{9}+\frac{21}{56}$

$\begin{array}{l}=\frac{5}{8}+\frac{3}{17}+\frac{2}{9}+\frac{-20}{17}+\frac{-2}{9}+\frac{3}{8}\\ =\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{3}{17}+\frac{-20}{17}\right)+\left(\frac{2}{9}+\frac{-2}{9}\right)\\ =1+\left(-1\right)+0=0\end{array}$

b) $\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}$

$\begin{array}{l}=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)\\ =-1+1=0\end{array}$

a) 

$\begin{array}{l}\frac{8}{19}+\frac{4}{21}+\frac{2}{5}+\frac{17}{21}+\frac{-27}{19}\\ =\left(\frac{8}{19}+\frac{-27}{19}\right)+\left(\frac{4}{21}+\frac{17}{21}\right)+\frac{2}{5}\\ =-1+1+\frac{2}{5}=\frac{2}{5}\end{array}$

b)

$\begin{array}{l}\frac{-5}{18}+\frac{7}{4}-\frac{-2}{3}+\frac{5}{6}\\ =\frac{-10}{36}+\frac{63}{36}+\frac{24}{36}+\frac{30}{36}\\ =\frac{107}{36}\end{array}$

a)

$\begin{array}{l}\frac{3}{8}-\frac{6}{18}+\frac{9}{27}+\frac{2}{16}\\ =\left(\frac{3}{8}+\frac{2}{16}\right)+\left(-\frac{6}{18}+\frac{9}{27}\right)\\ =\left(\frac{6}{16}+\frac{2}{16}\right)+\left(-\frac{18}{54}+\frac{18}{54}\right)\\ =\frac{1}{2}+0=\frac{1}{2}\end{array}$

b)

$\begin{array}{l}\frac{13}{26}-\frac{1}{3}-\frac{1}{2}+\frac{7}{21}\\ =\frac{1}{2}-\frac{1}{3}-\frac{1}{2}+\frac{1}{3}\\ =\left(\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)\\ =0+0=0\end{array}$

a) 

$\begin{array}{l}\frac{-17}{13}+\left(\frac{25}{101}+\frac{4}{13}\right)\\ =\frac{-17}{13}+\frac{25}{101}+\frac{4}{13}\\ =\left(\frac{-17}{13}+\frac{4}{13}\right)+\frac{25}{101}\\ =-1+\frac{25}{101}=\frac{-76}{101}\end{array}$

b)

$\begin{array}{l}\left(\frac{-5}{12}+\frac{6}{11}\right)+\left(\frac{7}{17}+\frac{5}{11}+\frac{5}{12}\right)\\ =\frac{-5}{12}+\frac{6}{11}+\frac{7}{17}+\frac{5}{11}+\frac{5}{12}\\ =\left(\frac{-5}{12}+\frac{5}{12}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{7}{17}\\ =0+1+\frac{7}{17}=1\frac{7}{17}\end{array}$

a)

$\begin{array}{l}\left(\frac{9}{16}+\frac{8}{-27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\\ =\frac{9}{16}+\frac{8}{-27}+1+\frac{7}{16}+\frac{-19}{27}\\ =\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{-8}{27}+\frac{-19}{27}\right)+1\\ =1+\left(-1\right)+1=1\end{array}$

b) 

$\begin{array}{l}\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{15}{16}-\frac{6}{15}\right)\\ =\frac{13}{5}+\frac{7}{16}-\frac{15}{16}+\frac{2}{5}\\ =\left(\frac{13}{5}+\frac{2}{5}\right)+\left(\frac{7}{16}-\frac{15}{16}\right)\\ =3+\frac{-1}{2}=\frac{5}{2}\end{array}$

a) 

$\begin{array}{l}-\left(\frac{3}{10}-\frac{6}{11}\right)-\left(\frac{21}{30}-\frac{5}{11}\right)\\ =-\frac{3}{10}+\frac{6}{11}-\frac{7}{10}+\frac{5}{11}\\ =\left(-\frac{3}{10}-\frac{7}{10}\right)+\left(\frac{6}{11}+\frac{5}{11}\right)\\ =-1+1=0\end{array}$

b) 

$\begin{array}{l}\left(\frac{13}{5}+\frac{7}{16}\right)-\left(\frac{11}{16}-\frac{12}{10}\right)\\ =\frac{13}{5}+\frac{7}{16}-\frac{11}{16}+\frac{6}{5}\\ =\left(\frac{13}{5}+\frac{6}{5}\right)+\left(\frac{7}{16}-\frac{11}{16}\right)\\ =\frac{19}{5}+\frac{-1}{4}=\frac{76}{20}+\frac{-5}{20}=\frac{71}{20}\end{array}$

$\begin{array}{l}\text{a) }x+\frac{2}{3}=\frac{-3}{4}\\ x=\frac{-3}{4}+\frac{2}{3}\\ x=\frac{-9}{12}+\frac{8}{12}\\ x=\frac{-1}{12}\end{array}$

$\begin{array}{l}\text{b) }\frac{5}{4}-x=\frac{-3}{2}\\ x=\frac{5}{4}-\frac{-3}{2}\\ x=\frac{5}{4}+\frac{6}{4}\\ x=\frac{11}{4}\end{array}$

a)

$\begin{array}{l}\frac{x}{2}-\frac{2}{5}=\frac{1}{10}\\ \frac{x}{2}=\frac{1}{10}+\frac{2}{5}\\ \frac{x}{2}=\frac{1}{10}+\frac{4}{10}\\ \frac{x}{2}=\frac{1}{2}\\ \Rightarrow x=1\end{array}$

b)

$\begin{array}{l}\frac{15}{x}-\frac{1}{3}=\frac{28}{51}\\ \frac{15}{x}=\frac{28}{51}+\frac{1}{3}\\ \frac{15}{x}=\frac{28}{51}+\frac{17}{51}\\ \frac{15}{x}=\frac{15}{17}\\ \Rightarrow x=17\end{array}$

a)

$\begin{array}{l}x-\frac{5}{12}-\frac{4}{9}=\frac{-13}{18}\\ x=\frac{-13}{18}+\frac{5}{12}+\frac{4}{9}\\ x=\frac{-26}{36}+\frac{15}{36}+\frac{16}{36}\\ x=\frac{5}{36}\end{array}$

b)

$\begin{array}{l}2-\left(x+\frac{3}{7}\right)=\frac{9}{-21}\\ 2-x-\frac{3}{7}=\frac{-3}{7}\\ -x=\frac{-3}{7}+\frac{3}{7}-2\\ -x=-2\\ \Rightarrow x=2\end{array}$

$\begin{array}{l}\left(\frac{-12}{27}+\frac{2}{3}\right)+\frac{-2}{9}\le x\le \left(\frac{11}{7}+\frac{2}{5}\right)+\frac{-7}{5}+\frac{3}{7}\\ \Rightarrow \frac{-4}{9}+\frac{-2}{9}+\frac{2}{3}\le x\le \frac{11}{7}+\frac{2}{5}+\frac{-7}{5}+\frac{3}{7}\\ \Rightarrow \frac{-2}{3}+\frac{2}{3}\le x\le \left(\frac{11}{7}+\frac{3}{7}\right)+\left(\frac{2}{5}+\frac{-7}{5}\right)\\ \Rightarrow 0\le x\le 2+\left(-1\right)\\ \Rightarrow 0\le x\le 1\\ \Rightarrow x=1\end{array}$

$\begin{array}{l}\left(\frac{-8}{13}+\frac{7}{17}\right)+\frac{21}{13}\le x\le \left(\frac{-9}{14}+3\right)+\frac{5}{-14}\\ \Rightarrow \frac{-8}{13}+\frac{7}{17}+\frac{21}{13}\le x\le \frac{-9}{14}+3+\frac{-5}{14}\\ \Rightarrow \left(\frac{-8}{13}+\frac{21}{13}\right)+\frac{7}{17}\le x\le \left(\frac{-9}{14}+\frac{-5}{14}\right)+3\\ \Rightarrow 1+\frac{7}{17}\le x\le -1+3\\ \Rightarrow \frac{24}{17}\le x\le 2\\ \Rightarrow 1,\mathrm{411...}\le x\le 2\\ \Rightarrow x=2\end{array}$

$\begin{array}{l}\frac{-3}{29}-\frac{2}{29}\le \frac{x}{29}<\frac{5}{29}-\frac{7}{29}.\\ \Rightarrow \frac{-5}{29}\le \frac{x}{29}<\frac{-2}{29}\\ \Rightarrow -5\le x<-2\\ \Rightarrow x\in \left\{-5;-4;-3\right\}\end{array}$

$\begin{array}{l}\frac{1}{6}-\frac{2}{5}<\frac{x}{30}\le \frac{1}{2}-\frac{8}{15}\\ \Rightarrow \frac{5}{30}-\frac{12}{30}<\frac{x}{30}\le \frac{15}{30}-\frac{16}{30}\\ \Rightarrow \frac{-7}{30}<\frac{x}{30}\le \frac{-1}{30}\\ \Rightarrow -7<x\le -1\\ \Rightarrow x\in \left\{-6;-5;-4;-3;-2;-1\right\}\end{array}$

$\begin{array}{l}\frac{-x}{2}+\frac{2x}{3}+\frac{x+1}{4}+\frac{2x+1}{6}=\frac{8}{3}\\ \Rightarrow \frac{-6x+8x+3\left(x+1\right)+2\left(2x+1\right)}{12}=\frac{32}{12}\\ \Rightarrow \frac{9x+5}{12}=\frac{32}{12}\\ \Rightarrow 9x+5=32\\ \Rightarrow x=3\end{array}$

$\begin{array}{l}\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\\ \Rightarrow \frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{12}{26}\\ \Rightarrow \frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{12}{26}\\ \Rightarrow \frac{3+5-2}{2x+1}=\frac{12}{26}\\ \Rightarrow \frac{6}{2x+1}=\frac{6}{13}\\ \Rightarrow 2x+1=13\\ \Rightarrow x=6\end{array}$

$\begin{array}{l}\text{B}=\text{B}=\frac{1}{3}+\left(\frac{1}{16}+\frac{1}{19}+\frac{1}{21}\right)+\left(\frac{1}{61}+\frac{1}{72}+\frac{1}{83}+\frac{1}{94}\right)\\ <\frac{1}{3}+\left(\frac{1}{15}+\frac{1}{15}+\frac{1}{15}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)\\ <\frac{1}{3}+\frac{3}{15}+\frac{4}{60}=\frac{5}{15}+\frac{3}{15}+\frac{1}{15}=\frac{3}{5}\\ \Rightarrow B<\frac{3}{5}\end{array}$

$\begin{array}{l}1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{2008.2009}\\ =1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{2008}-\frac{1}{2009}\\ =1+1-\frac{1}{2009}=1\frac{2008}{2009}\end{array}$

$\begin{array}{l}\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\\ =\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\\ =\frac{1}{3}-\frac{1}{87}=\frac{29}{87}-\frac{1}{87}=\frac{28}{87}\end{array}$

a) $\frac{6}{17}+\frac{3}{-17}=\frac{6}{17}+\frac{-3}{17}=\frac{3}{17}$

b) $\frac{-7}{18}+\frac{11}{-18}=\frac{-7}{18}+\frac{-11}{18}=-1$

c) $\frac{11}{6}+\frac{1}{4}=\frac{22+3}{12}=\frac{25}{12}$

a) $\frac{-3}{8}+\frac{2}{5}=\frac{-15+16}{40}=\frac{1}{40}$

b) $\frac{9}{7}-\frac{1}{5}=\frac{45-7}{35}=\frac{38}{35}$

c) $\frac{7}{11}-\left(-\frac{1}{3}\right)=\frac{7}{11}+\frac{1}{3}=\frac{21+11}{33}=1$

d) $\frac{-11}{33}-\left(\frac{1}{-4}\right)=\frac{-1}{3}+\frac{1}{4}=\frac{-4+3}{12}=\frac{-1}{12}$

a) $\frac{1}{5}+\frac{2}{9}+\frac{4}{5}+\frac{-11}{9}+\frac{3}{7}=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{2}{9}+\frac{-11}{9}\right)+\frac{3}{7}$$=1+(-1)+\frac{3}{7}=\frac{3}{7}$

b) $\frac{-7}{31}+\frac{24}{17}+\frac{7}{31}+\frac{-5}{17}+1=\left(\frac{-7}{31}+\frac{7}{31}\right)+\left(\frac{24}{17}+\frac{-5}{17}\right)+1$$=0+\frac{19}{17}+1=\frac{36}{17}$

a) $\frac{5}{16}-\frac{7}{13}-\frac{3}{16}+\frac{-6}{13}=\left(\frac{5}{16}-\frac{3}{16}\right)+\left(\frac{-7}{13}+\frac{-6}{13}\right)$$=\frac{1}{8}+\left(-1\right)=\frac{-7}{8}$

b) $\frac{-5}{12}-\left(\frac{3}{4}+\frac{7}{12}-\frac{2}{8}\right)=\frac{-5}{12}-\frac{3}{4}-\frac{7}{12}+\frac{2}{8}$$=\left(\frac{-5}{12}-\frac{7}{12}\right)+\left(-\frac{3}{4}+\frac{2}{8}\right)$$=-1+\left(\frac{-1}{2}\right)=\frac{-3}{2}$

a) $x=\frac{1}{36}+\frac{-3}{4}$

$\begin{array}{l}x=\frac{1+\left(-27\right)}{36}\\ x=\frac{-26}{36}\\ x=\frac{-13}{18}\end{array}$

b) $x=\frac{2}{5}+\frac{1}{4}+\frac{-1}{3}$

$\begin{array}{l}x=\frac{24+15+\left(-20\right)}{60}\\ x=\frac{19}{60}\end{array}$

a) $\frac{2}{5}+\frac{-3}{7}=\frac{x}{70}$

$\begin{array}{l}\frac{-1}{35}=\frac{x}{70}\\ \frac{-2}{70}=\frac{x}{70}\\ \Rightarrow x=-2\end{array}$

b) $\frac{5}{6}+\frac{-19}{30}=\frac{1}{x}$

$\begin{array}{l}\frac{1}{5}=\frac{1}{x}\\ \Rightarrow x=5\end{array}$

a) $x+\frac{1}{5}=\frac{-4}{9}$

$\begin{array}{l}x=\frac{-4}{9}-\frac{1}{5}\\ x=\frac{-29}{45}\end{array}$

b) $\frac{5}{7}-x=-\frac{9}{21}$

$\begin{array}{l}x=\frac{5}{7}+\frac{9}{21}\\ x=\frac{8}{7}\end{array}$

a) $x-\frac{-5}{6}=\frac{7}{12}+\frac{-1}{4}$

$\begin{array}{l}x=\frac{-5}{6}+\frac{7}{12}+\frac{-1}{4}\\ x=\frac{-1}{2}\end{array}$

b) $-x-\frac{1}{9}=-\frac{2}{45}$

$\begin{array}{l}x=\frac{-1}{9}+\frac{2}{45}\\ x=\frac{-1}{15}\end{array}$

a) $\frac{11}{3}-\frac{2}{3}\le x\le \frac{15}{9}-\left(-\frac{39}{9}\right)$

$\begin{array}{l}\iff 3\le x\le 6\\ \Rightarrow x\in \left\{3;4;5;6\right\}\end{array}$

b) $\frac{-5}{17}+\frac{3}{17}\le \frac{x}{17}\le \frac{13}{17}+\frac{-11}{17}$

$\begin{array}{l}\iff \frac{-2}{17}\le \frac{x}{17}\le \frac{2}{17}\\ \iff -2\le x\le 2\\ \Rightarrow x\in \left\{-2;-1;0;1;2\right\}\end{array}$

$\begin{array}{l}10A=\frac{{10}^{16}+10}{{10}^{16}+1}=1+\frac{9}{{10}^{16}+1}\\ 10B=\frac{{10}^{17}+10}{{10}^{17}+1}=1+\frac{9}{{10}^{17}+1}\end{array}$

Vì $\frac{9}{{10}^{16}+1}>\frac{9}{{10}^{17}+1}$ nên $10A>10B$

Vậy A > B

$\begin{array}{l}A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\mathrm{...}+\frac{1}{200}\\ =\left(\frac{1}{101}+\frac{1}{102}+\mathrm{...}+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+\mathrm{...}+\frac{1}{200}\right)\\ >\frac{1}{150}.50+\frac{1}{200}.50=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\end{array}$

$$\begin{gathered} S=\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{99.100} \\ \begin{array}{l}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{...}+\frac{1}{99}-\frac{1}{100}\\ =1-\frac{1}{100}=\frac{99}{100}\end{array} \end{gathered}$$