vãi 4 năm mà ko mtj thằng nào rep đã thế còn gấp

a) (2x.6) .(3x-18) = 0

=> 2x.6 = 0 => 12x = 0 => x = 0

3x - 18 = 0 => 3x = 18 => x = 6

KL:...

b) 25 + (15-x) = 30

25 + 15 - x = 30

40 - x = 30

x = 10

\(a)\left(2x.6\right).\left(3x-18\right)=0\Leftrightarrow\hept{\begin{cases}2x.6=0\\3x-18=0\end{cases}\Leftrightarrow\hept{\begin{cases}12x=0\\3x=18\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=6\end{cases}}}\)

\(b)25+\left(15-x\right)=30\Leftrightarrow15-x=5\Leftrightarrow x=20\)

tym cho mk nha

\(\left(2x.6\right)\left(3x-18\right)=0\)

<=> \(12x.\left(3x-18\right)=0\)

<=> \(\hept{\begin{cases}12x=0\\3x-8=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=0\\x=6\end{cases}}\)

\(25+\left(15-x\right)=30\)

<=> \(25+15-x=30\)

<=> \(x=0\)

học tốt

     \(x^4-2x^3-2x^2+3x+2=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x^3-2x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-x-x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^3-x\right)-\left(x+1\right)\right]=0\)

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

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

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

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

Đến đây ez r

=> |2x-1|=3x

=> 2x-1=3x hoặc 2x-1=-3x

=> x = -1 hoặc x = 1/5

k mk nha

                     =>    \(3x-15=0\) 

                     =>    \(3x=0+15\)

                     =>    \(3x=15\)

                     =>    \(x=15:3\)

                     =>    \(x=5\)

\(\left(3x-5\right)^7=0\)

\(\Rightarrow3x-5=0\)

\(\Rightarrow3x=5\)

\(\Rightarrow x=\frac{5}{3}\)

\(3x\left(2x-1\right)-1-2x=0\)

\(\Leftrightarrow6x^2-3x-1-2x=0\)

\(\Leftrightarrow6x^2-5x-1=0\)

chắc bài này sai đề. sửa lại:

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

\(\Leftrightarrow3x\left(2x-1\right)-\left(2x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{1}{2}\end{cases}}\)

a) 12+(2x-11)=53

(2x-11) = 53-12

2x-11= 41

2x=41+11

2x=52

x= 52:2

x=26

Vậy...

\(b,21-\left(-6+3x\right)=9\)

\(\Rightarrow21+6-3x=9\)

\(\Rightarrow27-3x=9\)

\(\Rightarrow3x=18\)

\(\Rightarrow x=6\)

c, -(2x+4)+11=-27

=>-2x-4+11=-27

=>-2x+7=-27

=>-2x = -34

=>x=17

d, 33-(33-x)=0

=>33-33+x=0

=>x=0

TH1: 2x-6=0                TH2: x-2=0

        2x=6                           x =2

        x=3

(2x-6).(x-2) = 0

=>2x - 6 = 0 hoặc x - 2 = 0

2x - 6 = 0

2x = 6

x = 3

x - 2 = 0

x = 2

Vậy x thuộc {2;3}