1) (3x + 9)(3x - 6) = 0

=> \(\orbr{\begin{cases}3x+9=0\\3x-6=0\end{cases}}\)

=> \(\orbr{\begin{cases}3x=-9\\3x=6\end{cases}}\)

=> \(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy ...

b) (2x + 15) - 25 = 47 - (10 - x)

=> 2x  - 10 = 37 + x

=> 2x - x = 37 + 10

=> x = 47

3, tương tự

4) |4 - 3x| = 8

=> \(\orbr{\begin{cases}4-3x=8\\4-3x=-8\end{cases}}\)

=> \(\orbr{\begin{cases}3x=-4\\3x=12\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{4}{3}\\x=4\end{cases}}\)

Vì x là số nguyên nên ...

còn lại tương tự

`123-x=38`

`=> x= 123-38`

`=>x=85`

Vậy `x=85`

__

`(x+15) -7=33`

`=>x+15=33+7`

`=>x+16= 40`

`=>x=40-16`

`=>x=24`

Vậy `x=24`

__

`3^(x+3) -13=230`

`=> 3^(x+3) = 230+13`

`=>3^(x+3)=243`

`=> 3^(x+3)=3^5`

`=> x+3=5`

`=>x=5-3`

`=>x=2`

Vậy `x=2`

a)        \(3x-15=25-5x\)

\(\Leftrightarrow\)\(3x+5x=25+15\)

\(\Leftrightarrow\)\(8x=40\)

\(\Leftrightarrow\)\(x=5\)

Vậy....

b)      \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow\)\(2x-17=-3x+18\)

\(\Leftrightarrow\)\(2x+3x=18+17\)

\(\Leftrightarrow\)\(5x=35\)

\(\Leftrightarrow\)\(x=7\)

Vậy.....

a) 3x-15=25-5x

3x+5x=25+15

8x=40

x=40/8

Vậy x=5

b)2x-17=-(3x-18)

2x-17=-3x+18

2x+3x=18+17

5x=35

x=35/5

Vậy x=7

b: \(\left(2x+1\right)^2=25\)

=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

c: \(\left(1-3x\right)^3=64\)

=>\(\left(1-3x\right)^3=4^3\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1

d: \(\left(4-x\right)^3=-27\)

=>\(\left(4-x\right)^3=\left(-3\right)^3\)

=>4-x=-3

=>x=4+3=7

e: \(x^2-5x=0\)

=>\(x\left(x-5\right)=0\)

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

a) 4.(18- 5x) - 12.( 3x-7) =15.(2x-16) - 6.(x+14)

72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

-20x - 36x - 30x + 6x = -240 - 84 - 72 -84 

-80x = -480

x= 6

4.(18 - 5x) - 12.(3x - 7) = 15.(2x - 16) - 6.(x+14)
4.18 - 4.5x - 12.3x + 12.7 = 15.2x - 15.16 - 6x - 6.14
72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
72 + 84 - 20x - 36x = 30x - 6x - 240 - 84
156 - 56x               = 24x - 324
156                        = 24x - 324 + 56x
156                       = 80x - 324
80x - 324               = 156
80x                       = 156 + 324
80x                       = 480
x                           = 480:80

x                           = 6
câu b giải tương tự  

a)   \(\left(2x-1\right)^2-25=0\)

⇔ \(\left(2x-1\right)^2-5^2=0\)

⇔  \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)

⇒  \(2x-1-5=0\) hoặc \(2x-1+5=0\)

⇔      \(x=3\)           hoặc  \(x=-2\)

Bài 1: Tìm x

a) (2x-1) ² - 25 = 0

<=> (2x-1)² =  25

<=>  2x-1 = 5  hay 2x-1 =-5

<=>  2x= 6      hay  2x=-4

<=>   x=3     hay    x= -2

Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0

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

<=> x-1=0  hay  3x+1=0

<=> x=1 hay 3x=-1

<=> x=1 hay x=\(\dfrac{-1}{3}\)

Vậy S={1;\(\dfrac{-1}{3}\)}

c) 2(x+3) - x ² - 3x = 0

<=> 2(x+3)- x(x+3)=0

<=> (x+3)(2-x)=0

<=> x+3=0 hay 2-x=0

<=> x=-3  hay  x=2

Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0

<=> x(x-2)+3(x-2)=0

<=> (x-2)(x+3)=0

<=> x-2=0 hay x+3=0

<=> x=2 hay x=-3

Vậy S={2;-3}
e) 4x ² - 4x +1 = 0

<=> (2x-1)²=0

<=> 2x-1=0

<=> 2x=1

<=> x=\(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}
f) x +5x²  = 0

<=> x(1+5x)=0

<=>x=0 hay 1+5x=0

<=> x=0 hay 5x=-1

<=> x=0 hay x= \(\dfrac{-1}{5}\)

Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0

<=> x²-x+3x-3=0

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

<=>  (x-1)(x+3)=0

<=> x-1=0 hay x+3=0

<=> x=1  hay x=-3

Vậy S={1;-3}

b) \(\text{3x (x-1) + x - 1 = 0}\)

\(\Rightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

c) \(\text{2(x+3) - x ² - 3x = 0}\)

\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Rightarrow\left(2-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d) \(\text{x(x - 2) + 3x - 6 = 0}\)

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

e)

\(\text{4x ² - 4x +1 = 0}\\ \Rightarrow\left(2x-1\right)^2=0\\ \Rightarrow2x-1=0\\ \Rightarrow x=0,5\)

f) \(\text{x +5x ² = 0}\)

\(\Rightarrow x\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

viết lại câu g đi bạn

giúp mik đi mà, gấp lắm, mik phải đi học ngay r, huhuhu

|x+1|+|3x-1|+|x-1|=3

=} vs cả trong dấu giá trị tuyệt đối >0 thì=}

x+1+3x-1+x-1=3{=}5x=4{=}x=4/5

=}vs cả trong giá trị tuyệt đối <0 thì=}

x+1+3x-1+x-1=-3{=}5x=-4{=}x=-4/5