Lỗi sai: Khi chuyển vế hạng tử -x từ vế phải sang vế trái và hạng tử -6 từ vế trái sang vế phải không đổi dấu của hạng tử đó.

Sửa lại:

3x – 6 + x = 9 – x

⇔ 3x + x + x = 9 + 6

⇔ 5x = 15

⇔ x = 3.

Vậy phương trình có nghiệm duy nhất x = 3.

     9(3x - 2) = x(2 - 3x)<=>27x - 18 = 2x - 3x²<=>3x² +27x - 2x = 18<=> 3x² +25x - 18= 0<=>3x² + 27x - 2x -18 = 0<=> 3x(x+9) - 2 (x+9) =0<=> (x+9) (3x-2) = 0\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) Vậy x=-9 hoặc x= \(\dfrac{2}{3}\)

\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(3x-2\right)\left(9+x\right)=0\)

\(\Rightarrow3x-2=0\) hoặc \(9-x=0\)

*) \(3x-2=0\)

\(3x=2\)

\(x=\dfrac{2}{3}\)

*) \(9+x=0\)

\(x=-9\)

Vậy \(x=\dfrac{2}{3}\); \(x=-9\)

\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(\Rightarrow9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(9+x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-9\end{matrix}\right.\)

9(3x-2)=x(2-3x)

9(3x-2)-x(2-3x)=0

9(3x-2)+x(3x-2)=0

(3x-2)(9+x)=0

=>TH1:3x-2=0;x=3/2

=>TH2:9+x=0;x=-9

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

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

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

\(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3.\)

\(\Rightarrow3x^2-6x-4x+8=3x^2-27x-3\)

\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Rightarrow17x=-11\)

\(\Leftrightarrow x=-\frac{11}{17}\)

(3x-4).(x-2)=3x(x-9)-3

<=> 3x^2-6x-4x+8=3x^2-27x-3

<=> 3x^2-6x-4x+8-3x^2+27x+3=0

<=> 17x+11=0

=> 17x=-11 => x=-11/17

\(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(< =>3x\left(x-2\right)-4\left(x-2\right)=3x\left(x-9\right)-3\)

\(< =>3x^2-10x+8=3x^2-27x-3\)

\(< =>3x^2-10x+8-3x^2+27x+3=0\)

\(< =>17x+11=0< =>17x=-11< =>x=-\frac{11}{17}\)

( 3x - 4 )( x - 2 ) = 3x( x - 9 ) - 3

⇔ 3x² - 10x + 8 = 3x² - 27x - 3

⇔ 3x² - 10x - 3x² + 27x = -3 - 8

⇔ 17x = -11

⇔ x = -11/17

Trả lời :

(3x - 4)(x - 2) = 3x(x -9) - 3

=> 3x² - 10x + 8 - 3x² + 27x + 3 = 0

=> 17x + 11 = 0

=> 17x = - 11

=> \(x=\frac{-11}{17}\)

\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

|3\(x\) - 1| +|1 - 3\(x\)| = 9

vì |3\(x\) - 1| = |1 - 3\(x\)| nên:

|3\(x\) - 1| + |1 - 3\(x\)| = |3\(x\) - 1| + |3\(x\) - 1| = 2|3\(\)\(x\) - 1|

⇒2.|3\(x\) - 1| = 9

      |3\(x\) - 1| = \(\dfrac{9}{2}\)

      \(\left[{}\begin{matrix}3x-1=\dfrac{-9}{2}\\3x-1=\dfrac{9}{2}\end{matrix}\right.\)

       \(\left[{}\begin{matrix}3x=-\dfrac{9}{2}+1\\3x=\dfrac{9}{2}+1\end{matrix}\right.\)

        \(\left[{}\begin{matrix}3x=-\dfrac{7}{2}\\3x=\dfrac{11}{2}\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=\dfrac{11}{6}\end{matrix}\right.\)

       Vậy \(x\) \(\in\) {- \(\dfrac{7}{6}\); \(\dfrac{11}{6}\)}