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

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.9=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+9x=27\\ 9x+27=27\\ 9x=0\\ x=0\)

a .   3 x = 9   ⇔ 3 x = 3 2 ⇔ x   =   2

b .   5 x = 5 3 ⇔ x = 3

c . 3 x + 1 = 3 2 ⇔ x + 1 = 2 ⇔ x = 1

a) Ta có: 3 x = 3 2  nên x = 2.

b) Ta có: 5 x = 5 3  nên x = 3.

c) Ta có:  3 x + 1   =   3 2 nên x +1 = 2, do đó x = 1.

x=0 nha bạn

|3x - 4| + |3x + 5| = 9

=> 3x. (-4 +5) = 9

=> 3x. 1 = 9

=> x = 3

Đoàn Thanh Bảo An làm kết quả đúng nhưng bn trình bày rõ ra mk tk cho

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}

\(x^3+3x^2+3x+9=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\x^2+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-3\\x^2=-3\Rightarrow ktm\end{cases}}\)

Vậy x=-3

ktm là gì vậy bạn

ktm là ko thỏa mãn

vì \(x^2\ge0\forall x\)

mà: x^2 =-3 

=> x^2 =-3 ko thỏa mãn

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

=> 3x² - 10x + 8 = 3x² - 27x - 3

=> 27x - 10x = -3 - 8

=> 17x = -11

=> x = -11/17 

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

\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Leftrightarrow17x+11=0\)

\(\Leftrightarrow17x=-11\)

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

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

\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Leftrightarrow27x-10x=-3-8\)

\(\Leftrightarrow x\left(27-10\right)=-11\)

\(\Leftrightarrow17x=-11\Leftrightarrow x=\frac{-11}{17}\)

(1-3x²)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)²

⇒1-3x²-(9x²+x-18x-2)=9x²-16-9(x²+6x+9)

⇒1-3x²-(9x²-17x-2)= -56x-97

⇒1-3x²-9x²+17x+2=-56x-97

⇒3-12x²+17x=-56x-97

⇒3-12x²+17x+56x+97=0

⇒-12x²+73x+100=0

⇒-(12x²-73x-100)=0