`\color{grey}\text{#071931}`

`(4 + 2x) * (9 - 3x) = 0`

TH1: `4 + 2x = 0 => 2x = -4 => x = -2`

TH2: `9 - 3x = 0 => 3x = 9 => x = 3`

Vậy, `x \in {-2; 3}`

____

`(3x - 19)^3 = 125`

`=> (3x - 19)^3 = 5^3`

`=> 3x - 19 = 5`

`=> 3x = 24`

`=> x = 8`

Vậy, `x = 8.`

mong các bn giúp mik, mik đag cần gấp 

1) |2x-1|=-19-x<=> \(\left[\begin{array}{nghiempt}2x-1=-19-x\\2x-1=19+x\end{array}\right.\)=> x=-6 hoặc x=20

2) |4-3x|=2x-10<=>\(\left[\begin{array}{nghiempt}4-3x=2x-10\\4-3x=10-2x\end{array}\right.\)=> x= 14/6 hoặc x=-6

3) |x|=3+2x<=> \(\left[\begin{array}{nghiempt}x=-3-2x\\x=3+2x\end{array}\right.\)=> x=-1 hoặc x=-3

1) - Nếu 2x - 1 < 0 thì -2x + 1 = -19 - x => -x = -20 => x = 20

    - Nếu 2x - 1 > 0 thì 2x - 1 = -19 - x => 3x = -18 => x = -6

2) - Nếu 4 - 3x < 0 thì -4 + 3x = 2x - 10 => 6 = -x => x = -6

    - Nếu 4 - 3x > 0 thì 4 - 3x = 2x - 10 => 14 = 5x => x = \(\frac{14}{5}\)

3) - Nếu x < 0 thì -x = 3 + 2x => -3x = 3 => x = -1

    - Nếu x > 0 thì x = 3 + 2x => -x = 3 => x = -3

1.

\(\left|2x-1\right|=-19-x\)

\(2x-1=\pm\left(-19-x\right)\)

TH1:

\(2x-1=-19-x\)

\(2x+x=-19-1\)

\(3x=-20\)

\(x=-\frac{20}{3}\)

TH2:

\(2x-1=19+x\)

\(2x-x=19-1\)

\(x=18\)

Vậy x = -20/3 hoặc x = 18

2.

\(\left|4-3x\right|=2x-10\)

\(4-3x=\pm\left(2x-10\right)\)

TH1:

\(4-3x=2x-10\)

\(-3x-2x=-10-4\)

\(-5x=-14\)

\(x=\frac{14}{5}\)

TH2:

\(4-3x=-2x+10\)

\(-3x+2x=10-4\)

\(x=-6\)

Vậy x = 14/5 hoặc x = -6

3.

\(\left|x\right|=3+2x\)

\(x=\pm\left(3+2x\right)\)

TH1:

\(x=3+2x\)

\(x-2x=3\)

\(x=-3\)

TH2:

\(x=-3-2x\)

\(x+2x=-3\)

\(3x=-3\)

\(x=-1\)

1) | 2x-1| = -19 -x

Th1:

2x -1 = -19 -x

3x = -18

x= -6

Th2:

2x -1 = -(-19 -x)

2x -1 = 19 +x

x= 20

Vậy...

2) | 4- 3x| = 2x- 10

Th1:

4 - 3x = 2x -10

5x = 14

x= 14/5

Th2:

4- 3x = -(2x-10)

4 - 3x = -2x +10

x= -6

Vậy...

3) | x| = 3+ 2x

Th1:

3 + 2x = x

x= -3

Th2:

3 + 2x = -x

3x = -3 

x= -1

Vậy...

1) không có x

2) x=14 , x=-6

3)x=-3 hoặc x= -1

1)2x-1=-(-19-x)

2x-1=19+x

2x-x=19+1

x=20

1) \(\left|2x-1\right|=-19-x\)

\(=>\orbr{\begin{cases}-19-x=2x-1\\-19-x=-\left(2x-1\right)\end{cases}=>\orbr{\begin{cases}-19+1=2x+x\\-19-x=-2x+1\end{cases}}}\)

                                                                 \(=>\orbr{\begin{cases}-18=3x\\-x+2x=1+19\end{cases}=>\orbr{\begin{cases}x=-6\\x=20\end{cases}}}\)     

2) \(\left|4-3x\right|=2x-10\)   

\(=>\orbr{\begin{cases}2x-10=4-3x\\2x-10=-\left(4-3x\right)\end{cases}=>\orbr{\begin{cases}2x+3x=4+10\\2x-10=-4+3x\end{cases}}}\)    

                                                               \(=>\orbr{\begin{cases}5x=14\\-10+4=3x-2x\end{cases}=>\orbr{\begin{cases}x=\frac{14}{5}\\x=-6\end{cases}}}\)     

3) \(\left|x\right|=3+2x\)     

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

                                            \(=>\orbr{\begin{cases}x=-3\\3x=-3\end{cases}=>\orbr{\begin{cases}x=-3\\x=-1\end{cases}}}\)           

Ủng hộ mk nha ^_- 

a) \(PT\Leftrightarrow3x-2x=2-3\Leftrightarrow x=-1\)

Vậy: \(S=\left\{-1\right\}\)

b) \(PT\Leftrightarrow-2x+3x=-7+22\Leftrightarrow x=15\)

Vậy: \(S=\left\{15\right\}\)

c) \(PT\Leftrightarrow8x-5x=3+12\Leftrightarrow3x=15\Leftrightarrow x=5\)

Vậy: \(S=\left\{5\right\}\)

d) \(PT\Leftrightarrow x+4x-2x=12+25-1\Leftrightarrow3x=36\Leftrightarrow x=12\)

Vậy: \(S=\left\{12\right\}\)

e) \(PT\Leftrightarrow x+2x+3x-3x=19+5\Leftrightarrow3x=24\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=12

=>x=4

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

a)3x-2=2x-3

=>x=-1

b)7-2x=22-3x

=>x=15

c)8x-3=5x+12

=>3x=15

=>x=5

d)x-12+4x=25+2x-1

=>3x=36

=>x=12

e)x+2x+3x-19=3x+5

=>3x=24

=>x=8

A,

a: \(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)

=>-6x-5=0

=>-6x=5

hay x=-5/6

b: \(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)

=>8x+16=0

hay x=-2

c: \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)

=>9x-10=0

hay x=10/9

d: \(\Leftrightarrow10x-15-20x+28=19-2x^2-4x-2\)

\(\Leftrightarrow-10x+13+2x^2+4x-17=0\)

\(\Leftrightarrow2x^2-6x-4=0\)

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

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)