Bài tập: Cộng, trừ số hữu tỉ

$\frac{25}{12}+\frac{-4}{12}=\frac{25+\left(-4\right)}{12}=\frac{21}{12}=\frac{7}{4}$

$-\frac{4}{7}-\frac{2}{-3}=-\frac{4}{7}+\frac{2}{3}=\frac{-12}{21}+\frac{14}{21}=\frac{2}{21}$

$\frac{-10}{8}+\frac{15}{4}=\frac{-5}{4}+\frac{15}{4}=\frac{-5+15}{4}=\frac{10}{4}=\frac{5}{2}$

$3\frac{1}{4}-1\frac{1}{3}=\frac{13}{4}-\frac{4}{3}=\frac{39}{12}-\frac{16}{12}=\frac{23}{12}$

$\frac{3}{8}+\frac{-14}{6}=\frac{9}{24}+\frac{-56}{24}=\frac{9+\left(-56\right)}{24}=\frac{-47}{24}$

$\begin{array}{l}\frac{5454}{5757}-\frac{171717}{191919}=\frac{54.101}{57.101}-\frac{17.10101}{19.10101}\\ =\frac{54}{57}-\frac{17}{19}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\end{array}$

$\begin{array}{l}\left[\frac{5}{8}+\left(\frac{-3}{4}\right)\right]+\frac{15}{6}=\frac{15}{24}+\frac{-18}{24}+\frac{60}{24}\\ =\frac{15+\left(-18\right)+60}{24}=\frac{57}{24}=\frac{19}{8}\end{array}$

$\begin{array}{l}\frac{17}{2}-\left(\frac{-3}{7}+\frac{5}{3}\right)=\frac{17}{2}-\left(\frac{-9}{21}+\frac{35}{21}\right)\\ =\frac{17}{2}-\frac{26}{21}=\frac{357}{42}-\frac{52}{42}=\frac{305}{42}\end{array}$

$\frac{1}{2}-\left(\frac{3}{4}+\frac{-5}{6}\right)-\frac{7}{12}=\frac{1}{2}-\frac{3}{4}+\frac{5}{6}-\frac{7}{12}=\frac{6-9+10-7}{12}=0$

$\begin{array}{l}-3-\frac{2}{3}+\left(\frac{-10}{9}-\frac{25}{3}\right)-\frac{5}{6}\\ =-3-\frac{2}{3}-\frac{10}{9}-\frac{25}{3}-\frac{5}{6}\\ =\frac{-54-12-20-150-15}{18}=\frac{-251}{18}\end{array}$

$\begin{array}{l}\left(\frac{5}{-7}-\frac{-5}{-7}\right)+\frac{4}{3}\\ =\left(\frac{-5}{7}-\frac{5}{7}\right)+\frac{4}{3}\\ =\frac{-10}{7}+\frac{4}{3}\\ =\frac{-30+28}{21}\\ =\frac{-2}{21}\end{array}$

$\begin{array}{l}\frac{7}{3}+\left[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)\right]\\ =\frac{7}{3}+\frac{-5}{6}+\frac{-2}{3}\\ =\frac{7}{3}+\frac{-2}{3}+\frac{-5}{6}\\ =\frac{5}{3}+\frac{-5}{6}\\ =\frac{10+\left(-5\right)}{6}\\ =\frac{5}{6}\end{array}$

$\begin{array}{l}\frac{4}{3}-\left[\left(\frac{-11}{6}\right)-\left(\frac{2}{9}+\frac{5}{3}\right)\right]\\ =\frac{4}{3}+\frac{11}{6}+\left(\frac{2}{9}+\frac{5}{3}\right)\\ =\frac{4}{3}+\frac{11}{6}+\frac{2}{9}+\frac{5}{3}\\ =\frac{24+33+4+30}{18}\\ =\frac{91}{18}\end{array}$

$$\begin{gathered} \left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right) \\ \begin{array}{l}=8-\frac{9}{4}+\frac{2}{7}+6+\frac{3}{7}-\frac{5}{4}-3-\frac{2}{4}+\frac{9}{7}\\ =\left(8+6-3\right)+\left(-\frac{9}{4}-\frac{5}{4}-\frac{2}{4}\right)+\left(\frac{2}{7}+\frac{3}{7}+\frac{9}{7}\right)\\ =11-4+2\\ =9\end{array} \end{gathered}$$

$\begin{array}{l}A=\frac{1}{3}-\frac{3}{4}-\left(-\frac{3}{5}\right)+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\\ =\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\\ =\left(\frac{1}{3}+\frac{3}{5}+\frac{1}{15}\right)+\left(-\frac{3}{4}-\frac{2}{9}-\frac{1}{36}\right)+\frac{1}{72}\\ =\left(\frac{5}{15}+\frac{9}{15}+\frac{1}{15}\right)+\left(-\frac{27}{36}-\frac{8}{36}-\frac{1}{36}\right)+\frac{1}{72}\\ =1-1+\frac{1}{72}\\ =\frac{1}{72}\\ \end{array}$

$\begin{array}{l}B=\frac{3}{99.96}-\frac{3}{96.93}-\frac{3}{93.90}-\mathrm{...}-\frac{3}{7.4}-\frac{3}{4}\\ =\frac{3}{99.96}-\left(\frac{3}{96.93}+\frac{3}{96.93}+\mathrm{...}+\frac{3}{7.4}+\frac{3}{4.1}\right)\\ =\frac{3}{99.96}-\left(\frac{3}{1.4}+\frac{3}{4.7}+\mathrm{...}+\frac{3}{90.93}+\frac{3}{93.96}\right)\\ =\frac{3}{99.96}-\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\mathrm{...}+\frac{1}{90}-\frac{1}{93}+\frac{1}{93}-\frac{1}{96}\right)\\ =\left(\frac{1}{96}-\frac{1}{99}\right)-\left(1-\frac{1}{96}\right)\\ =\frac{1}{96}-\frac{1}{99}-1+\frac{1}{96}\\ =\frac{1}{48}-1-\frac{1}{99}\\ =-\frac{47}{48}-\frac{1}{99}=\frac{-1567}{1584}\end{array}$

$$\begin{gathered} x+\frac{1}{5}=\frac{3}{7} \\ \begin{array}{l}x=\frac{3}{7}-\frac{1}{5}\\ x=\frac{15-7}{35}\\ x=\frac{8}{35}\end{array} \end{gathered}$$

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

$$\begin{gathered} \frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3} \\ \begin{array}{l}\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\\ \frac{2}{5}+x=\frac{11}{12}-\frac{8}{12}\\ \frac{2}{5}+x=\frac{1}{4}\end{array} \end{gathered}$$

$\begin{array}{l}x=\frac{1}{4}-\frac{2}{5}\\ x=\frac{-3}{20}\end{array}$

$$\begin{gathered} x+\frac{2}{3}=\frac{3}{5}-\left(-\frac{1}{6}\right) \\ \begin{array}{l}x+\frac{2}{3}=\frac{3}{5}+\frac{1}{6}\\ x+\frac{2}{3}=\frac{23}{30}\\ x=\frac{23}{30}-\frac{2}{3}\\ \begin{array}{l}x=\frac{23-20}{30}\\ x=\frac{1}{10}\end{array}\end{array} \end{gathered}$$

$\begin{array}{l}x-\left[\frac{17}{2}-\left(\frac{-3}{7}+\frac{5}{3}\right)\right]=\frac{-1}{3}\\ \begin{array}{l}x-\left(\frac{17}{2}+\frac{3}{7}-\frac{5}{3}\right)=\frac{-1}{3}\\ x-\left(\frac{357}{42}+\frac{18}{42}-\frac{70}{42}\right)=\frac{-1}{3}\\ x-\frac{305}{42}=\frac{-1}{3}\\ x = \frac{-1}{3}+\frac{305}{42}\\ x=\frac{-14}{42}+\frac{305}{42}\\ x=\frac{291}{42}=\frac{97}{14}\end{array}\end{array}$

$\frac{9}{2}-\left[\frac{2}{3}-\left(x+\frac{7}{4}\right)\right]=\frac{-5}{4}$

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

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

$\frac{2}{3}-\left(x+\frac{7}{4}\right)=\frac{23}{4}$

$x+\frac{7}{4}=\frac{2}{3}-\frac{23}{4}$

$x+\frac{7}{4}=\frac{8-69}{12}$

$x+\frac{7}{4}=\frac{-61}{12}$

$x=\frac{-61}{12}-\frac{7}{4}$

$x=\frac{-61}{12}-\frac{21}{12}$

$x=\frac{-82}{12}$

$x=\frac{-41}{6}.$

Vậy $x=\frac{-41}{6}.$

$$\begin{gathered} \frac{x}{2}-\frac{1}{x}=\frac{1}{12} \\ \begin{array}{l}\frac{x^2-2}{2x}=\frac{1}{12}\\ 12\left(x^2-2\right)=2x\\ 6\left(x^2-2\right)=x\\ 6x^2-x-12=0\\ \begin{array}{l}6x^2+8x-9x-12=0\\ 2x\left(3x+4\right)-3\left(3x+4\right)=0\\ \left(3x+4\right)\left(2x-3\right)=0\end{array}\end{array} \end{gathered}$$

TH1: 3x + 4 = 0

$\begin{array}{l}3x=-4\\ x=-\frac{4}{3}\end{array}$

TH2: 2x - 3 = 0

$$\begin{gathered} \begin{array}{l}2x=3\\ x=\frac{3}{2}\end{array} \\ Vậy x\in \left\{\frac{-4}{3};\frac{3}{2}\right\} \end{gathered}$$

$\left(2x-\frac{1}{2}\right).\left(3x-\frac{1}{3}\right)<0$

TH1: $\left\{\begin{array}{l}2x-\frac{1}{2}>0\\ 3x-\frac{1}{3}<0\end{array}\right.\iff \left\{\begin{array}{l}2x>\frac{1}{2}\\ 3x<\frac{1}{3}\end{array}\right.\iff \left\{\begin{array}{l}x>\frac{1}{4}\\ x<\frac{1}{9}\end{array}\right.\Rightarrow$không tìm được x

TH2: $\left\{\begin{array}{l}2x-\frac{1}{2}<0\\ 3x-\frac{1}{3}>0\end{array}\right.\iff \left\{\begin{array}{l}2x<\frac{1}{2}\\ 3x>\frac{1}{3}\end{array}\right.\iff \left\{\begin{array}{l}x<\frac{1}{4}\\ x>\frac{1}{9}\end{array}\right.\iff \frac{1}{9}<x<\frac{1}{4}$

Vậy: $\frac{1}{9}<x<\frac{1}{4}$

$\frac{x+\frac{3}{2}}{x-\frac{2}{3}}<0 Đk: x\ne \frac{2}{3}$

Ta có: $x-\frac{2}{3}<x+\frac{3}{2}$ nên

$\left\{\begin{array}{l}x-\frac{2}{3}<0\\ x+\frac{3}{2}>0\end{array}\right.\iff \left\{\begin{array}{l}x<\frac{2}{3}\\ x>-\frac{3}{2}\end{array}\right.\iff -\frac{3}{2}<x<\frac{2}{3}$

Vậy: $-\frac{3}{2}<x<\frac{2}{3}$

Ta có: $4<x<\frac{21}{5}$

Vì $\frac{21}{5}=4,2$ nên 4 < x < 4,2 

Vậy không có số nguyên x nào thỏa mãn yêu cầu bài toán.

Ta có: 0 < x < 2

Vì x nguyên nên x = 1

Vậy x = 1.

$$\begin{gathered} Vì \frac{-9}{2}<x\le -4 và x nguyên \\ Nên x = -4 \end{gathered}$$

$$\begin{gathered} Vì -\frac{11}{9}<x<0 \\ Nên x=-1 (do x nguyên) \end{gathered}$$

$$\begin{gathered} \frac{1}{x}-\frac{y}{6}=\frac{1}{3} \\ \begin{array}{l}\frac{6-xy}{6x}=\frac{1}{3}\\ 3\left(6-xy\right)=6x\\ 6-xy=2x\\ xy+2x=6\\ x\left(y+2\right)=6\end{array} \end{gathered}$$

Vì x, y là các số nguyên mà $6=6.1=\left(-6\right).\left(-1\right)=2.3=\left(-2\right).\left(-3\right)$

Ta có:

Vậy  $\left(x;y\right)=\left(-6;-3\right);\left(-3;-4\right);\left(-2;-5\right);\left(-1;-8\right);\left(1;4\right);\left(2;1\right);\left(3;0\right);\left(6;-1\right).$

$$\begin{gathered} \frac{x}{2}+\frac{3}{y}=\frac{5}{4} \\ \begin{array}{l}\frac{xy+6}{2y}=\frac{5}{4}\\ 4\left(xy+6\right)=10y\\ 2xy+12=5y\\ 2xy-5y=-12\\ y\left(2x-5\right)=-12\end{array} \end{gathered}$$

Vì x,y nguyên nên 2x-5 nguyên

mà $-12=1.\left(-12\right)=12.\left(-1\right)=2.\left(-6\right)=6.\left(-2\right)=3.\left(-4\right)=4.\left(-3\right)$

nên

(các ước khác của 12 không thỏa mãn 2x - 5 nguyên)

Vậy $\left(x;y\right)=\left(-1;4\right);\left(2;12\right);\left(3;-12\right);\left(4;-4\right)$

$\frac{-1}{21}+\frac{-1}{14}=\frac{-2}{42}+\frac{-3}{42}=-\frac{5}{42}$

$\frac{-1}{9}-\frac{5}{12}=-\frac{19}{36}$

$\frac{-14}{20}+0,6=-\frac{1}{10}$

$4,5-\left(-\frac{7}{5}\right)=\frac{59}{10}$

Ba cách viết như sau:

$\frac{-4}{15}=\frac{-1}{15}+\frac{-1}{5};\frac{-4}{15}=\frac{-1}{30}+\frac{-7}{30};\frac{-4}{15}=\frac{-2}{15}+\frac{-2}{15}$

Ta thực hiện $\frac{-1}{12}+\frac{-5}{6}-\frac{4}{3}=\frac{-2}{24}+\frac{-20}{24}+\frac{-32}{24}=\frac{-54}{24}=\frac{-9}{4}$

Ta thực hiện $\left(\frac{-24}{11}+\frac{2}{11}\right)+\left(\frac{-19}{13}+\frac{-20}{13}\right)=(-2)+(-3)=-5$

Ta thực hiện $-x=\frac{4}{5}-\frac{3}{10}-\frac{16}{5}=\frac{-27}{10}=>x=\frac{27}{10}$

$\begin{array}{l}x-\frac{8}{5}=\frac{1}{20}-\frac{1}{10}=>x-\frac{8}{5}=\frac{-1}{20}\\ =>x=\frac{-1}{20}+\frac{8}{5}=>x=\frac{31}{20}\end{array}$

$\frac{-11}{25}=\frac{-1}{5}+\frac{-6}{25};\frac{-11}{25}=\frac{-3}{25}+\frac{-8}{25};\frac{-11}{25}=\frac{-2}{25}+\frac{-9}{25}$

$\frac{-11}{25}=\frac{4}{25}-\frac{3}{5};\frac{-11}{25}=\frac{1}{25}-\frac{12}{25};\frac{-11}{25}=\frac{3}{2}-\frac{97}{50}$

$$\begin{gathered} A=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{15} \\ =>A=\frac{13}{15}. \end{gathered}$$

Ta có:

 $$\begin{gathered} B=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{7.8}+\frac{1}{8.9}\right) \\ = \frac{1}{9.10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right) \\ = \frac{1}{90}-\left(1-\frac{1}{9}\right) \\ =\frac{1}{90}-\frac{8}{9}=\frac{1}{90}-\frac{80}{90}=-\frac{79}{90} \end{gathered}$$

Vậy B = $-\frac{79}{90}$