a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

`-4/5 + 2x =1/3 -2/3x`

`=>  2x + 2/3x =1/3 +4/5`

`=> (2 +2/3) x= 5/15 + 12/15`

`=> ( 6/3 +2/3) x= 17/15`

`=> 8/3 x= 17/15`

`=> x= 17/15 : 8/3`

`=> x= 17/15 xx 3/8`

`=>x=17/40`

` @ ` \(yeuugialinhh\)

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

\(< =>2x+3=-x-2\)

\(< =>2x+x=-2-3\)

\(< =>3x=-5< =>x=-\frac{5}{3}\)

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

\(\Rightarrow2x+3=-x-2\)

\(\Rightarrow2x-\left(-x\right)=-2-3\)

\(\Rightarrow3x=-5\)

\(\Rightarrow x=-\frac{5}{3}\)

2x+3=-(x+2)

2x+3=-x+2

2x+x=-2-3

2x+x=-5

x×(2+1)=-5

x×3=-5

x=-5:3

x=-5/3

vậy x=-5/3

\(\Leftrightarrow\left(x-2021\right)\left(x-5\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\)

1: Ta có: \(20-2\left(x+4\right)=4\)

\(\Leftrightarrow2\left(x+4\right)=16\)

\(\Leftrightarrow x+4=8\)

hay x=4

5: Ta có: \(\left(x+1\right)^3=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

a  A=4x-x^2+3

      =(x-2)^2-1

     MIN A= -1 khi (x-2)^2=0

      x-2=0

      x=2

B=x-x^2

B=-x^2+x

-B=x^2-x

-B=(x-1/2)^2-1/4

B=-(x-1/2)^2+1/4

MAX B=1/4 khi -(x-1/2)^2=0

x-1/2=0

x=1/2

N=2x-2x^2-5

-N=2x^2-2x+5

-N=2(x^2-x+2)+1

-N=2{(x-1/2)^2+7/4}+1

-N=2(x-1/2)^2+7/2+1

-N=2(x-1/2)^2+9/2

N=-2(x-1/2)^2-9/2

MAX N=-9/2 khi -2(x-1/2)^2=0

x-1/2=0

x=1/2

Câu a) của bạn là 6 nhân 4 hay \(6,4\) vậy bạn? Nguyễn Thanh Giang

`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12