1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2

3)
(2x+1)²=5²
2x+1=5
2x=4
x=2
 

\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)

\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

a) Ta có: \(x\left(x-1\right)-x^2+2x=5\)

\(\Leftrightarrow x^2-x-x^2+2x=5\)

hay x=5

b) Ta có: \(2x^2-2x=\left(x-1\right)^2\)

\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c) Ta có: \(\left(x+3\right)\cdot\left(x^2-3x+9\right)-x\left(x-2\right)^2=19\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-19=0\)

\(\Leftrightarrow x^3+8-x^3+4x^2-4x=0\)

\(\Leftrightarrow4x^2-4x+8=0\)(Vô lý)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

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