\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)

\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)

\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)

\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)

\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)

\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)

= ...

Tự tính

Bài 2 : \(a,3-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}\)

\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}+3\)

\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{2}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{2}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{23}{3}\\x=\frac{-19}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{23}{3};\frac{-19}{3}\right\}\)

b, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)

\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)

\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)

\(\Rightarrow0,6-1,6< x\le3\)

\(\Rightarrow-1< x\le3\)

\(\Rightarrow x\in\left\{0;1;2;3\right\}\)

Tham khảo link  :   https://hoidap247.com/cau-hoi/884527

:))))

1.8100

2. 34 x 18 + 18 x 66

= 18 x ( 34 + 66)

= 18 x 100 = 1800

3. X × ( 8 +  12) =  160 + 20 × 12

= X x 20 = 160 + 240

= X x 20 = 400

X = 400 : 20 = 20

4. X x 12 - X x 2 = 2020

(12 - 2) x X = 2020

10 x X = 2020

X = 2020 : 10 = 202

thank bạn nha 

lỗi đề

a) \(14:\frac{0,4x+0,6}{x}=7\)

\(\frac{0,4x+0,6}{x}=2\)

0,4x + 0,6 = 2.x

2x - 0,4x = 0,6

1,6x = 0,6

x = 0,375

b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)

\(\left(160\%+\frac{2}{3}x-x\right)=55\)

\(x\left(\frac{2}{3}-1\right)=53,4\)

\(-\frac{1}{3}x=\frac{267}{5}\)

\(x=\frac{267}{5}.\frac{3}{-1}\)

\(x=-160,2\)

c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)

\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)

\(1:\frac{1}{2^{36}}=2^x\)

\(2^{36}=2^x\)

\(x=36\)

Câu 1 : \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}=\frac{x+2}{21}+\frac{x+2}{22}\)
      => \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}-\frac{x+2}{21}-\frac{x+2}{22}=0\)
      =>  x+2 . ( \(\frac{1}{18}+\frac{1}{19}+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\)) = 0
     Vì  \(\frac{1}{18}+\frac{1}{19}_{ }+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\ne0\)nên  x+2=0 
                                                                              => x= 0 - 2 = -2
                       Vậy x = -2