a) Ta có:

$\frac{3}{4}+\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{3}{4}+\left(\frac{3}{6}-\frac{2}{6}\right)=\frac{3}{4}+\frac{1}{6}$

$=\frac{18}{24}+\frac{4}{24}=\frac{22}{24}=\frac{11}{12}$

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

$=\frac{15}{12}-\frac{4}{12}=\frac{15-4}{12}=\frac{11}{12}$

Vậy $\frac{3}{4}+\left(\frac{1}{2}-\frac{1}{3}\right)$ do kết quả đều bằng $\frac{11}{12}.$

b) Ta có:

$\frac{2}{3}-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{2}{3}-\left(\frac{3}{6}+\frac{2}{6}\right)$

$=\frac{2}{3}-\frac{5}{6}=\frac{4}{6}-\frac{5}{6}=\frac{-1}{6}$

$\frac{2}{3}-\frac{1}{2}-\frac{1}{3}=\frac{4}{6}-\frac{3}{6}-\frac{2}{6}$

$=\frac{1}{6}-\frac{2}{6}=\frac{-1}{6}$

Vậy $\frac{2}{3}-\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{2}{3}-\frac{1}{2}-\frac{1}{3}$ do kết quả đều bằng $\frac{-1}{6}.$

$A=\left(7-\frac{2}{5}+\frac{1}{3}\right)-\left(6-\frac{4}{3}+\frac{6}{5}\right)-\left(2-\frac{8}{5}+\frac{5}{3}\right)$

$A=7-\frac{2}{5}+\frac{1}{3}-6+\frac{4}{3}-\frac{6}{5}-2+\frac{8}{5}-\frac{5}{3}$

$A=\left(7-6-2\right)+\left(\frac{-2}{5}-\frac{6}{5}+\frac{8}{5}\right)+\left(\frac{1}{3}+\frac{4}{3}-\frac{5}{3}\right)$

A = -1+ 0 + 0 = -1

$x-\frac{2}{5}=\frac{1}{2}$

a) 

$$\begin{gathered} x+\frac{1}{2}=-\frac{1}{3} \\ x=-\frac{1}{3}-\frac{1}{2} \\ x=\frac{-2}{6}-\frac{3}{6} \\ x=\frac{-5}{6} \end{gathered}$$

Vậy $x=\frac{-5}{6}$

b) 

$$\begin{gathered} \left(\frac{-2}{7}\right)+x=-\frac{1}{4} \\ x=\frac{-1}{4}-\frac{-2}{7} \\ x=\frac{-7}{28}-\frac{-8}{28} \\ x=\frac{1}{28} \end{gathered}$$

Vậy $x=\frac{1}{28}$

$$\begin{gathered} a) 1\frac{1}{2}+\frac{1}{5}.\left[\left(-2\frac{5}{6}\right)+\frac{1}{3}\right] \\ =\frac{3}{2}+\frac{1}{5}.\left[\frac{-17}{6}+\frac{2}{6}\right] \\ =\frac{3}{2}+\frac{1}{5}.\frac{-15}{6} \\ =\frac{3}{2}+\frac{\left(-1\right)\mathrm{.3.5}}{\mathrm{5.3.2}}=\frac{3}{2}-\frac{1}{2}=1 \end{gathered}$$

$$\begin{gathered} b) \frac{1}{3}.\left(\frac{2}{5}-\frac{1}{2}\right):{\left(\frac{1}{6}-\frac{1}{5}\right)}^2 \\ =\frac{1}{3}.\left(\frac{4}{10}-\frac{5}{10}\right):{\left(\frac{5}{30}-\frac{6}{30}\right)}^2 \\ =\frac{1}{3}.\frac{-1}{10}:{\left(\frac{1}{30}\right)}^2 \\ =\frac{-1}{30}.{\left(\frac{30}{1}\right)}^2=\frac{-1}{30}.\frac{900}{1}=-30 \end{gathered}$$

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

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

$=-1+\frac{5}{6}=\frac{-6}{6}+\frac{5}{6}=\frac{-1}{6}$

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

$=\left(\frac{3}{5}-\frac{1}{5}\right)-\frac{2}{3}=\frac{2}{5}-\frac{2}{3}$

$=\frac{6}{15}-\frac{10}{15}=\frac{-4}{15}$

c) $\left[\left(\frac{-1}{3}\right)+1\right]-\left(\frac{2}{3}-\frac{1}{5}\right)=\frac{-1}{3}+1-\frac{2}{3}+\frac{1}{5}$

$=\left(\frac{-1}{3}-\frac{2}{3}\right)+1+\frac{1}{5}=\frac{-3}{3}+1+\frac{1}{5}$

$=-1+1+\frac{1}{5}=\frac{1}{5}$

d) $1\frac{1}{3}+\left(\frac{2}{3}-\frac{3}{4}\right)-\left(0,8+1\frac{1}{5}\right)$

$=\frac{4}{3}+\frac{2}{3}-\frac{3}{4}-0,8-1\frac{1}{5}$

$=\frac{4}{3}+\frac{2}{3}-\frac{3}{4}-\frac{4}{5}-\frac{6}{5}$

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

$=\frac{6}{3}-\frac{10}{5}-\frac{3}{4}=2-2-\frac{3}{4}=\frac{3}{4}$

a) $\left(\frac{3}{4}:1\frac{1}{2}\right)-\left(\frac{5}{6}:\frac{1}{3}\right)=\left(\frac{3}{4}:\frac{3}{2}\right)-\left(\frac{5}{6}.\frac{3}{1}\right)$

$=\left(\frac{3}{4}.\frac{2}{3}\right)-\left(\frac{5}{6}.\frac{3}{1}\right)=\frac{1}{2}-\frac{5}{2}=\frac{-4}{2}=-2$

b) $\left[\left(\frac{-1}{5}\right):\frac{1}{10}\right]-\frac{5}{7}.\left(\frac{2}{3}-\frac{1}{5}\right)$

$=\frac{-1}{5}.\frac{10}{1}-\frac{5}{7}.\left(\frac{10}{15}-\frac{3}{15}\right)$

$=-2-\frac{5}{7}.\frac{7}{15}=-2-\frac{1}{3}$

$=\frac{-6}{3}-\frac{1}{3}=\frac{-7}{3}$

c) $\left(-0,4\right)+2\frac{2}{5}.{\left[\left(\frac{-2}{3}\right)+\frac{1}{2}\right]}^2$

$=\frac{-2}{5}+\frac{12}{5}.{\left[\frac{-4}{6}+\frac{3}{6}\right]}^2$

$=\frac{-2}{5}+\frac{12}{5}.{\left[\frac{1}{6}\right]}^2=\frac{-2}{4}+\frac{12}{5}.\frac{1}{36}$

$=\frac{-2}{5}+\frac{1}{15}=\frac{-6}{15}+\frac{1}{15}=\frac{-5}{15}=\frac{-1}{3}$

d) $\left\{\left[{\left(\frac{1}{25}-0,6\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\left[\left(\frac{-1}{3}\right)+\frac{1}{2}\right]$

$=\left\{\left[{\left(\frac{1}{25}-\frac{3}{5}\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\left[\frac{-2}{6}+\frac{3}{6}\right]$

$=\left\{\left[{\left(\frac{1}{25}-\frac{3}{5}\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\left[\frac{-2}{6}+\frac{3}{6}\right]$

$=\left\{\left[{\left(\frac{1}{25}-\frac{15}{25}\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\frac{1}{6}$

$=\left\{\left[{\left(\frac{-14}{25}\right)}^2:\frac{49}{125}\right].\frac{5}{6}\right\}-\frac{1}{6}$

$=\left\{\left[\frac{{\left(-2\right)}^2{.7}^2}{{\left(5^2\right)}^2}:\frac{7^2}{5^3}\right].\frac{5}{6}\right\}-\frac{1}{6}$

$=\left\{\left[\frac{{\left(-2\right)}^2{.7}^2}{5^4}.\frac{5^3}{7^2}\right].\frac{5}{6}\right\}-\frac{1}{6}$

$=\left\{\frac{{\left(-2\right)}^2{.7}^2}{5^4}.\frac{5^3}{7^2}.\frac{5}{6}\right\}-\frac{1}{6}$

$=\frac{4}{6}-\frac{1}{6}$

$=\frac{3}{6}=\frac{1}{2}$

 

a) $x+\frac{3}{5}=\frac{2}{3}$

$$\begin{gathered} x=\frac{2}{3}-\frac{3}{5} \\ x=\frac{10}{15}-\frac{9}{15} \\ x=\frac{1}{15} \end{gathered}$$

Vậy $x=\frac{1}{15}$

b) $\frac{3}{7}-x=\frac{2}{5}$

$$\begin{gathered} x=\frac{3}{7}-\frac{2}{5} \\ x=\frac{15}{35}-\frac{14}{35} \\ x=\frac{1}{35} \end{gathered}$$

Vậy $x=\frac{1}{35}$

c) $\frac{4}{9}-\frac{2}{3}x=\frac{1}{3}$

$$\begin{gathered} \frac{2}{3}x=\frac{4}{9}-\frac{1}{3} \\ \frac{2}{3}x=\frac{4}{9}-\frac{3}{9} \\ \frac{2}{3}x=\frac{1}{9} \\ x=\frac{1}{9}:\frac{2}{3} \\ x=\frac{1}{9}.\frac{3}{2} \\ x=\frac{1}{6} \end{gathered}$$

Vậy $x=\frac{1}{6}$

d) $\frac{3}{10}x-1\frac{1}{2}=\left(\frac{-2}{7}\right):\frac{5}{14}$

$$\begin{gathered} \frac{3}{10}x-\frac{3}{2}=\frac{-2}{7}.\frac{14}{5} \\ \frac{3}{10}x-\frac{3}{2}=\frac{-4}{5} \\ \frac{3}{10}x=\frac{-4}{5}+\frac{3}{2} \\ \frac{3}{10}x=\frac{-8}{10}+\frac{15}{10} \\ \frac{3}{10}x=\frac{7}{10} \\ x=\frac{7}{10}:\frac{3}{10} \\ x=\frac{7}{10}.\frac{10}{3} \\ x=\frac{7}{3} \end{gathered}$$

Vậy$x=\frac{7}{3}$

a) $\frac{2}{9}:x+\frac{5}{6}=0,5$

$\frac{2}{9}:x=\frac{1}{2}-\frac{5}{6}$

$\frac{2}{9}:x=\frac{3}{6}-\frac{5}{6}$

$\frac{2}{9}:x=\frac{-2}{6}$

$x=\frac{2}{9}:\frac{-2}{6}$

$x=\frac{2}{9}.\frac{6}{-2}$

$x=\frac{-2}{3}$

Vậy $x=\frac{-2}{3}$

b) $\frac{3}{4}-\left(x-\frac{2}{3}\right)=1\frac{1}{3}$

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

$x-\frac{2}{3}=\frac{3}{4}-\frac{4}{3}$

$x-\frac{2}{3}=\frac{9}{12}-\frac{16}{12}$

$x-\frac{2}{3}=\frac{-7}{12}$

$x=\frac{-7}{12}+\frac{2}{3}$

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

$x=\frac{1}{12}$

Vậy $x=\frac{1}{12}$

c) $1\frac{1}{4}:\left(x-\frac{2}{3}\right)=0,75$

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

$x-\frac{2}{3}=\frac{5}{4}:\frac{3}{4}$

$x-\frac{2}{3}=\frac{5}{4}.\frac{4}{3}$

$x-\frac{2}{3}=\frac{5}{3}$

$x=\frac{5}{3}+\frac{2}{3}$

$x=\frac{7}{3}$

d) $\left(-\frac{5}{6}x+\frac{5}{4}\right):\frac{3}{2}=\frac{4}{3}$

$-\frac{5}{6}x+\frac{5}{4}=\frac{4}{3}.\frac{3}{2}$

$-\frac{5}{6}x+\frac{5}{4}=2$

$-\frac{5}{6}x=2-\frac{5}{4}$

$-\frac{5}{6}x=\frac{8}{4}-\frac{5}{4}$

$-\frac{5}{6}x=\frac{3}{4}$

$x=\frac{3}{4}:\left(-\frac{5}{6}\right)$

$x=\frac{3}{4}.\frac{-6}{5}$

$x=\frac{-9}{10}$

Vậy $x=\frac{-9}{10}$

 

a) $\frac{13}{23}.\frac{7}{11}+\frac{10}{23}.\frac{7}{11}$

$=\frac{7}{11}.\left(\frac{13}{23}+\frac{10}{23}\right)$

$=\frac{7}{11}.\frac{23}{23}=\frac{7}{11}.1=\frac{7}{11}$

b) $\frac{5}{9}.\frac{23}{11}-\frac{1}{11}.\frac{5}{9}+\frac{5}{9}$

$=\frac{5}{9}.\left(\frac{23}{11}-\frac{1}{11}+1\right)$

$=\frac{5}{9}.\left(\frac{22}{11}+1\right)$

$=\frac{5}{9}.\left(2+1\right)$

$=\frac{5}{9}.3=\frac{5}{3}$

c) $\left[\left(-\frac{4}{9}\right)+\frac{3}{5}\right]:\frac{13}{17}+\left(\frac{2}{5}-\frac{5}{9}\right):\frac{13}{17}$

$=\left[\left(-\frac{4}{9}\right)+\frac{3}{5}\right].\frac{17}{13}+\left(\frac{2}{5}-\frac{5}{9}\right).\frac{17}{13}$

$=\left(-\frac{4}{9}+\frac{3}{5}+\frac{2}{5}-\frac{5}{9}\right).\frac{17}{13}$

$=\left(\frac{-4-5}{9}+\frac{3+2}{5}\right).\frac{17}{13}$

$=\left(\frac{-9}{9}+\frac{5}{5}\right).\frac{17}{13}$

$=\left(-1+1\right).\frac{17}{13}$

$=0.\frac{17}{13}=0$

d) $\frac{3}{16}:\left(\frac{3}{22}-\frac{3}{11}\right)+\frac{3}{16}:\left(\frac{1}{10}-\frac{2}{5}\right)$

$=\frac{3}{16}:\left(\frac{3}{22}-\frac{6}{22}\right)+\frac{3}{16}:\left(\frac{1}{10}-\frac{4}{10}\right)$

$=\frac{3}{16}:\frac{-3}{22}+\frac{3}{16}:\frac{-3}{10}$

$=\frac{3}{16}.\frac{22}{-3}+\frac{3}{16}.\frac{10}{-3}$

$=\frac{3}{16}.\left(\frac{22}{-3}+\frac{10}{-3}\right)$

$=\frac{3}{16}.\frac{32}{-3}$

$=\frac{\mathrm{3.16.2}}{\mathrm{16.3.}(-1)}=-2$