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dễ thương quá minion 

d, X/8*32+X/3*18-X/5*10=120

   X*32/8+X*18/3-X*10/5=120

  32X/8+18X/3-10X/5=120

  4X+6X-2X=120

  8X=120

 X=120/8

X=15

a, ( 25 * 12 - X - 25 ) : 25 = 4

( 25 * 12 - x - 25 )           = 4 * 25 

( 25 * 12 - x - 25 )           = 100

300 - x - 25                    = 100

300 - x                           = 100 + 25

300 - x                           = 125

x                                   = 300 - 125

x                                   = 175

a) <=> 15-5x-20=-12-3

<=> -5x=-12-3-15+20=-10

=>x=-10:(-5)=2

b)<=>7-x-25-7=-25

<=> -x=-25-7+25+7=0 =>x=0

c) /x+2/=0 => x+2=0 =>x=-2

d) sai đề

e)<=> /x-5/ = 7

<=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)

g) <=> -x-20-8+2x=-15

<=> x=-15+20+8=13

d là | x + 3 | = 7 - ( - 2 )

(-25)-(5.x-3)=4

5.x-3 = -25 - 4

5.x-3=-29

5.x=-26

x=-26/5

x=-5,2

5.x-6=20-(12-6)

5.x-6=20-6

5.x-6=14

5.x=20

x=20:5

x=4

vậy x=4

1/
-25-(5x-3)=4
-25-4=5x-3
-25-4+3=5x
5x= -26
x= -26/5
2/
5x-6=20-(12-6)
5x = 20-12+6+6
5x = 20
x = 20/5 = 4
3/
-2x -40 = (5-x) -(-15+60
-2x -40 = 5-x +15-60
-2x +x = 5+15-60+40
-x = 0
x=0

1234567

_Mấy bác cứ thik đăng nhiều :v , nhìn mak ko muốn lm . E lm bài 1 thôi :v còn các bài còn lại bác tự lm ( nó cx dễ thôi mà ) _

Bài 1 :

\(a) 2x-13=25+6x\)

\(\Rightarrow2x-6x=25+13\)

\(\Rightarrow-4x=38\)

\(\Rightarrow x=-\dfrac{19}{2}\)

Vậy .......

\(b) 12-x=3x+6\)

\(\Rightarrow-x-3x=6-12\)

\(\Rightarrow-4x=-6\)

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

Vậy .....

\(c) 40-(25-2x)=x\)

\(\Rightarrow40-25+2x=x\)

\(\Rightarrow15+2x=x\)

\(\Rightarrow2x-x=-15\)

\(\Rightarrow x=-15\)

Vậy ......

\(d) |x-3|=5\)

\(\Rightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

Vậy ....

e) \(|x-3|+(x+2)+(x+1)=12\)

\(\Rightarrow\left|x-3\right|+x+2+x+1=12\)

\(\Rightarrow\left|x-3\right|+2x+3=12\)

\(\Rightarrow\left|x-13\right|+2x=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3+2x=9\\-\left(x-3\right)+2x=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\) \(( x = 6 \) ko thỏa mãn điều kiện )

Vậy ....

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 \(x\in\left\{5;-5;-10\right\}.\)

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

`#040911`

`a)`

`6 \times x - 5 = 613`

`=> 6 \times x = 613 + 5`

`=> 6 \times x = 618`

`=> x = 618 \div 6`

`=> x = 103`

Vậy, `x = 103`

`b)`

`12 \times x + 3 \times x = 30`

`=> x \times (12 + 3) = 30`

`=> x \times 15 = 30`

`=> x = 30 \div 15`

`=> x = 2`

Vậy, `x = 2`

`c)`

`125 - 25 \times (x - 1) = 100`

`=> 25 \times (x - 1) = 125 - 100`

`=> 25 \times (x - 1) = 25`

`=> x - 1 = 25 \div 25`

`=> x - 1 = 1`

`=> x = 1 + 1`

`=> x = 2`

Vậy, `x = 2`

`d)`

`(x - 2) \times (9x - 4) = 0?`

`=>`

TH1: `x - 2 = 0`

`=> x = 0 + 2`

`=> x = 2`

TH2: `9x - 4 = 0`

`=> 9x = 4`

`=> x = 4/9`

Vậy, `x \in {2; 4/9}.`

\(a,6x-5=613\\ \Leftrightarrow6x=618\\ \Leftrightarrow x=103\\ b,12x+3x=30\\ \Leftrightarrow15x=30\\ \Leftrightarrow x=2\\ c,125-25\left(x-1\right)=100\\ \Leftrightarrow25\left(x-1\right)=25\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,\left(x-2\right)\cdot\left(9x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{9}\end{matrix}\right.\)