a)

\(\begin{array}{l}x + \frac{1}{2} =  - \frac{1}{3}\\x =  - \frac{1}{3} - \frac{1}{2}\\x =  - \frac{2}{6} - \frac{3}{6}\\x = \frac{{ - 5}}{6}\end{array}\)     

Vậy \(x = \frac{{ - 5}}{6}\).

 b)

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

Vậy \(x = \frac{1}{{28}}\).

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

 a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)

\(\begin{array}{l}x:1\frac{2}{7} =  - 3,5\\x:\frac{9}{7} =  - \frac{7}{2}\\x =  - \frac{7}{2}.\frac{9}{7}\\x =  - \frac{9}{2}\end{array}\)

b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)

\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

#)Giải :

1. 

Ta có : \(\frac{n+1}{n+2}>\frac{n}{n+2}>\frac{n}{n+3}\)

\(\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)

2. 

a) \(x\left(104,5-14,1+9,6\right)=25\)

\(x\times100=25\)

\(x=25\div100\)

\(x=0,25\)

Bài 1 : Ta có :\(\frac{n+1}{n+2}>\frac{n}{n+2}>\frac{n}{n+3}\)

\(\Leftrightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)

Bài 2 : \(104,5\cdot x-14,1\cdot x+9,6\cdot x=25\)

\(\Leftrightarrow\left[104,5-14,1+9,6\right]\cdot x=25\)

\(\Leftrightarrow100\cdot x=25\)

\(\Leftrightarrow x=\frac{1}{4}\)

\(1+2+3+4+...+x=210\)

Số số hạng của dãy là : \((x-1):1+1=x\) số

Cho nên tổng của dãy đó là : \(\frac{x(x+1)}{2}=210\)

\(\Leftrightarrow x(x+1)=420\)

\(\Leftrightarrow x(x+1)=20\cdot21\)

\(\Leftrightarrow x=20\)

\(x-\frac{3}{4}=1-\frac{5}{6}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{6}+\frac{3}{4}=\frac{11}{12}\)

\(a,2^x=8\\ \Leftrightarrow2^x=2^3\\ \Leftrightarrow x=3\\ b,2^x=\dfrac{1}{4}\\ \Leftrightarrow2^x=2^{-2}\\ \Leftrightarrow x=-2\\ c,2^x=\sqrt{2}\\ \Leftrightarrow2^x=2^{\dfrac{1}{2}}\\ \Leftrightarrow x=\dfrac{1}{2}\)

m hay lắm Dương, t gửi câu hỏi, m cũng gửi!!! Good Job

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

2x -5 - 9x -\(\frac{3}{2}\)            =4

2x - 9x                    = 4+ 5+ \(\frac{3}{2}\)

B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)

\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)

\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)

\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))

mik fan Phong ca nè bạn