3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow2x-\dfrac{3}{4}\ge0\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)

Vậy xảy ra khi:

\(x=\dfrac{3}{2}\)

mk giải cho bạn ở trên r đó

12x/6 + 4x/6 - 3x/6 = 3/8 - 1/4 - 1/6

13x/6 = (9 - 6 - 4)/24 = -1/24

--> x = -1/52

|x-3|+1=x

|x+3| = x-1

Th1: x+3 = x-1 

0x = 4

x thuộc rỗng

Th2: -(x+3) = x-1

-x -3 = x-1

2x = -2

x= -1 

Vậy x=-1

|x-3|=2x+4

Th1: x-3 = 2x +4

x = -7

Th2: -(x-3) = 2x+4

-x +3 = 2x +4

-3x = 1

x= -1/3 

Vậy x= -7; x= -1/3

|3-2x|=x+1

Th1: 3 - 2x = x+1 

3x = 2

x= 2/3

Th2: - (3-2x) = x+1

- 3 +2x = x+1

x= 4

Vậy x= 2/3 và x= 4

|x-1|+3x=1

|x+1| = 1- 3x

Th1: x+1 = 1- 3x 

4x = 0

x=0 

Th2: -(x+1) = 1- 3x

-x -1 = 1- 3x

2x = 2

x=1

Vậy x=0 và x=1

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