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

\(\Leftrightarrow6x^2-3x-1-2x=0\)

\(\Leftrightarrow6x^2-5x-1=0\)

chắc bài này sai đề. sửa lại:

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

\(\Leftrightarrow3x\left(2x-1\right)-\left(2x-1\right)=0\)

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

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

x<-2 hoac

x>-1/3

\(\left(2x-2\right).\left(3x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-2=0\\3x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=2\\3x=-6\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)

ĐS: ...........

( 2x - 2 ) . ( 3x + 6 ) = 0

 => 2x - 2 = 0 hoặc 3x + 6 = 0

2x = 0 + 2 hoặc 3x = 0 - 6

2x = 2 hoặc 3x = - 6

x = 1 hoặc x = - 2

*Lưu ý nếu bạn học số nguyên rồi thì mới làm theo cách này nha!

II3x-3I+2x+1I=3x+2021^0

II3x-3I+2x+1I=3x+1

\(\)ĐK:3x+1\(\ge\)0

3x\(\ge\)-1

x\(\ge\frac{-1}{3}\)

\(\Rightarrow\)I3x-3I+2x+1=3x+1

I3x-3I=x

\(\Rightarrow\)3x-3=\(\pm\)x

TH1:3x-3=x                     TH2:3x-3=-x

2x=3                                       4x=3

x=\(\frac{3}{2}\)                                    x=\(\frac{3}{4}\)

Vậy x=\(\frac{3}{2}\); x=\(\frac{3}{4}\)

thanks

ta co (2x-1)(3x+1)+(3x+4)(3-2x)=5

     (=)6x²-3x+2x-1+6x-6x²+12-8x=5

     (=)-4x+11=5

     (=)-4x=-6

     (=)x=3/2

(2x-1)(3x+1)+(3x-4)(3-2x)=5

<=> 6x²+2x-3x-1+9x-6x²-12+8x=5

<=> 16x-13=5

<=> 16x = 18

<=> x=9/8

     \(x^4-2x^3-2x^2+3x+2=0\)

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

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

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

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

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

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

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

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

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

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

Đến đây ez r