Ta có: \(\left(\dfrac{2}{3}x-\dfrac{4}{9}\right)\left(\dfrac{1}{2}-\dfrac{3}{7}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{4}{9}\\\dfrac{3}{7}x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{7}{6}\end{matrix}\right.\)

Ta có: \(\dfrac{x+1}{2018}+\dfrac{x+1}{2019}+\dfrac{x+1}{2020}+\dfrac{x+1}{2021}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

a) \(\left|x+\dfrac{4}{9}\right|-\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Rightarrow\left|x+\dfrac{4}{9}\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\)

b) \(\left|x-\dfrac{4}{11}\right|+\left|5+y\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

\(a,\Rightarrow\left|x+\dfrac{4}{9}\right|=\dfrac{3}{2}+\dfrac{1}{2}=2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{9}=2\\x+\dfrac{4}{9}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{14}{9}\\x=-\dfrac{22}{9}\end{matrix}\right.\\ b,\Rightarrow\left\{{}\begin{matrix}x-\dfrac{4}{11}=0\\5+y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{11}\\y=-5\end{matrix}\right.\)

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

\(\left|x+1\right|-4x=0\\ \Leftrightarrow\left|x+1\right|=4x\)

Ta có : \(\left\{{}\begin{matrix}x+1\ge0\Leftrightarrow x\ge-1\\x+1< 0\Leftrightarrow x< -1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4x\\\left(-x+1\right)=4x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-4x=-1\\-x-1=4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x=-1\\-x-4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=-\dfrac{1}{5}\left(tm\right)\end{matrix}\right.\)

`@ Kidd`

\(\left|x+1\right|-4x=0\)

\(\Leftrightarrow\left|x+1\right|=4x\left(1\right)\)

\(\text{Ta có:}\left\{{}\begin{matrix}x+1\ge0\text{ hay }x\ge-1\\x+1< 0\text{ hay }x< -1\end{matrix}\right.\)

\(TH1:x+1=4x\)

\(\Leftrightarrow x-4x=-1\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\left(\text{nhận}\right)\)

\(TH2:-\left(x+1\right)=4x\)

\(\Leftrightarrow-x-1=4x\)

\(\Leftrightarrow-x-4x=1\)

\(\Leftrightarrow-5x=1\)

\(\Leftrightarrow x=\dfrac{-1}{5}\left(\text{nhận}\right)\)

\(\text{Vậy phương trình (1) có tập nghiệm là }S=\left\{\dfrac{1}{3};\dfrac{-1}{5}\right\}\)

Ta có : x³+x²-x-1

=(x³+x²)-(1x+1)

=x²(x+1)-1(x+1)

=(x²-1)(x+1)

=(x-1)(x+1)(x+1)

=(x-1)(x+1)²

= (x+1)(x² - 1) = 0

th1: x+1 =0 

x = -1

th2: x² -1 =0 

x = +_ 1

kl: x=1; x = -1