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 ^_- 

(x-2)²-(x-3)(x-3)=6

x²-2.x.2+2²-x²-3²=6

(x²-x²)-4x+(2²⁺3²)=6

-4x+13=6

-4x=6-13=-7

x=-7:(-4)=1,75

a/Ta có: M(x)+N(x) = (2x⁵ - 4x³ + 2x² + 10x - 1) + (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10)

                              = 2x⁵ - 2x⁵ - 4x³ + 4x³ + 2x⁴ + 2x² + x² + 10x + x -1 - 10

                              = 2x⁴ + 3x² + 11x - 11

b/ Ta có: A(x) = N(x)-M(x) = (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10) - (2x⁵ - 4x³ + 2x² + 10x - 1)

                                         = -2x⁵ - 2x⁵ + 2x⁴ + 4x³ + 4x³ + x² - 2x² + x - 10x -10 + 1

                                         = -2x⁵ + 2x⁴ + 8x³ - x² - 9x -9

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`