\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)

a) Ta có: \(x^2-x-12=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b) Ta có: \(x^2+3x-18=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=3\end{matrix}\right.\)

a: \(3x\left(x-3\right)+4x-12=0\)

=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)

=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(3x+4\right)=0\)

=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)

\(\Leftrightarrow x^3+1-x^3+2x=17\)

=>2x+1=17

=>2x=17-1=16

=>\(x=\dfrac{16}{2}=8\)

c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)

=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)

=>\(15x=-14\)

=>\(x=-\dfrac{14}{15}\)

a)

\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)

c)

\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

d)

\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

e)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)

f)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)

\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{2}{5}\)

\(x=\dfrac{5}{10}-\dfrac{4}{10}\)

\(\Rightarrow x=....\)

\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{5-4}{10}=\dfrac{1}{10}\)

\(b,x-\dfrac{2}{5}=\dfrac{1}{7}\\ \Rightarrow x=\dfrac{1}{7}+\dfrac{2}{5}=\dfrac{5+14}{35}=\dfrac{19}{35}\)

\(c,x\cdot\dfrac{3}{4}=\dfrac{9}{20}\\ \Rightarrow x=\dfrac{9}{20}:\dfrac{3}{4}=\dfrac{9}{20}\cdot\dfrac{4}{3}=\dfrac{3\cdot1}{5\cdot1}=\dfrac{3}{5}\)

\(d,x:\dfrac{1}{7}=14\\ \Rightarrow x=14\cdot\dfrac{1}{7}=\dfrac{14}{7}=2\)

\(e,\dfrac{2}{3}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{2}{3}-\dfrac{1}{5}=\dfrac{10-3}{15}=\dfrac{7}{15}\)

\(f,\dfrac{4}{15}:x=\dfrac{12}{25}\\ \Rightarrow x=\dfrac{4}{15}:\dfrac{12}{25}=\dfrac{4}{15}\cdot\dfrac{25}{12}=\dfrac{1\cdot5}{3\cdot3}=\dfrac{5}{9}\)

a, \(x+\dfrac{2}{5}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}-\dfrac{2}{5}\\ \Rightarrow x=\dfrac{5}{10}-\dfrac{4}{10}\\ \Rightarrow x=\dfrac{1}{10}\)

b) \(x-\dfrac{2}{5}=\dfrac{1}{7}\\ \Rightarrow x=\dfrac{1}{7}+\dfrac{2}{5}\\ \Rightarrow x=\dfrac{5}{35}+\dfrac{14}{35}\\ \Rightarrow x=\dfrac{19}{35}\)

c) \(x\times\dfrac{3}{4}=\dfrac{9}{20}\\ \Rightarrow x=\dfrac{9}{20}:\dfrac{3}{4}\\ \Rightarrow x=\dfrac{9}{20}\times\dfrac{4}{3}\\ \Rightarrow x=\dfrac{36}{60}\\ \Rightarrow x=\dfrac{3}{5}\)

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

\(2^{x+2}\cdot3^{x+1}\cdot5x=10800\)

\(2^x\cdot2^2\cdot3^x\cdot2\cdot5^x=10800\)

\(\left(2\cdot3\cdot5\right)^x\cdot12=10800\)

\(30^x=10800\div12\)

\(30^x=900\)

\(\Leftrightarrow30^{x=}=30^2\Leftrightarrow x=2\)

chả hiểu

bạn nguyễn thu thủy làm sai r nhé 
10800:2=5400
chứ không phải 18000:2=900

`a,`

`x*3+7=16`

`=>3x = 16 - 7`

`=> 3x = 9`

`=> x = 9 \div 3`

`=> x = 3`

Vậy, `x = 3`

`b,`

` x- 152 \div 2 = 46?`

`=> x - 76 = 46`

`=> x = 46 + 76`

`=> x = 122`

Vậy, `x = 122.`

`c,`

` 74-2*(x+3)=34`

`=> 2(x + 3) = 74 - 34`

`=> 2(x+2) = 40`

`=> x + 2 = 40 \div 2`

`=> x + 2 = 20`

`=> x = 20 - 2`

`=> x = 18`

Vậy, `x = 18.`