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

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

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

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

\(B1:\)

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

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

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

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

\(x^2-6x^2+12x-8=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

       (3x - 1 )² - 16 =0

<=>(3x - 1 )² = 16

<=>3x-1  =4

<=>x=5/3

(3x - 1)² - 16 = 0

<=> (3x - 1)² - 4² = 0

<=> (3x - 1 - 4)(3x - 1 + 4) = 0

<=> (3x - 5)(3x + 3)  = 0

<=> 3x - 5 = 0 hoặc 3x + 3 = 0

<=> x = 5/3 hoặc x = - 1

(3x-1)^2-16=0

(3x-1)^2=16  

Th1: 3x-1=4

        3x=5  

        x=5/3

Th2: 3x-1=-4    

        3x=-3            

          x=-1 

\(3x^3+2x^2+2x+3=0\)

\(\Leftrightarrow3\left(x^3+1\right)+2x\left(x+1\right)=0\)

\(\Leftrightarrow3\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)=0\)

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

Mà \(3x^2-x+3=3\left[\left(x-\frac{1}{6}\right)^2+\frac{35}{36}\right]>0\forall x\)

Do đó: \(x+1=0\Leftrightarrow x=-1\)

Tập nghiệm: \(S=\left\{-1\right\}\)

\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)

\(\Leftrightarrow\left[\left(x-1\right)+\left(2x+3\right)\right]\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=27x^3+8\)

\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(6x^2-15x-9\right)=0\)(Chuyển vế)

\(\Leftrightarrow3\left(3x+2\right)\left(2x^2-5x-3\right)=0\)

\(\Leftrightarrow3\left(3x+2\right)\left(x-3\right)\left(2x+1\right)=0\)

Tập nghiệm: \(S=\left\{-\frac{2}{3};3;-\frac{1}{2}\right\}\)

a, 3x^2-27x=0

3x(x-9)=0

3x=0=>x=0

x-9=0=>x=9

b,2/3x(x^2-4)=0

2/3x=0=>x=0

x^2-4=0=>x=2

\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-9=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

\(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

\(\left[\begin{array}{nghiempt}x=0\\x-2=0\\x+2=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=0\\x=2\\x=-2\end{array}\right.\)

a)\(3x^2-27x=0\)

\(3x\left(x-9\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}3x=0\\x-9=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=9\end{array}\right.\)

b) \(\frac{2}{3}x\left(x^2-4\right)=0\)

\(\frac{2}{3}x\left(x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}\frac{2}{3}x=0\\x+2=0\\x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=2\end{array}\right.\)

Ta có: \(x^2-27x-64=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{27}{2}+\frac{729}{4}-\frac{985}{4}=0\)

\(\Leftrightarrow\left(x-\frac{27}{2}\right)^2=\frac{985}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{27}{2}=\sqrt{\frac{985}{4}}\\x-\frac{27}{2}=-\sqrt{\frac{985}{4}}\end{matrix}\right.\)\(\left[{}\begin{matrix}x=\sqrt{\frac{985}{4}}+\frac{27}{2}=\frac{27+\sqrt{985}}{2}\\x=-\sqrt{\frac{985}{4}}+\frac{27}{2}=\frac{27-\sqrt{985}}{2}\end{matrix}\right.\)

Vậy: \(x=\frac{27\pm\sqrt{985}}{2}\)

a,  3(x+3)-2(x-5)=11

=> 3x+9-2x+10=11

=> 3x-2x=11-10-9

=>  x=-8

Vậy.........

b,   14-4|x|=-6

=>  -4|x|=8

=>   |x|=-2(VL vì trị tuyệt đối luôn lớn hơn hoặc = 0)

Vậy......