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`a,(2x-5)(12+5x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

`b, (x-3)(x-4)-2(x-3)=0`

`<=>(x-3)(x-4-2)=0`

`<=>(x-3)(x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

`c, x(x-1)(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

`d, (2x)/3 +(2x-1)/6=0`

`<=> (4x)/6 +(2x-1)/6=0`

`<=> (4x+2x-1)/6=0`

`<=> (6x-1)/6=0`

`<=> 6x-1=0`

`<=> 6x=1`

`<=>x=1/6` ( đề là vậy à bạn )

 a) \(\left(2x-5\right)\left(12+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\\x=0\end{matrix}\right.\)

d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)

\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)

\(\Leftrightarrow6x-1=0\)

\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)

a) \(x-2=0\Leftrightarrow x=2\)

b) \(x^2-2x=0\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

e) \(2x^2+5x+3=0\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\end{matrix}\right.\)

f) \(x^2-x-12=0\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

b)     \(x^3-6x^2+11x-12=0\)

\(\Leftrightarrow\)\(x^3-4x^2-2x^2+8x+3x-12=0\)

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

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

Ta có:    \(x^2-2x+3\)

        \(=\left(x-1\right)^2+2>0\)

\(\Rightarrow\)\(x-4=0\)

\(\Leftrightarrow\)\(x=4\)

Vậy...

a)x=4-12=-8

b)x=19-0=19

c)x=-12+4:2=-4

d)x=6+(-2):2=2

2x+69.2=69.4

2x+138=276

2x = 276-138

2x = 138

x = 138:2

x = 69

2x-12-x = 0

<=> 2x-x-12 =0

(2x-x)-12=0

=> x-12=0

x = 0+12

x  =12

(x-7)(2x-8)=0

\(\Rightarrow\hept{\begin{cases}x-7=0\\2x-8=0\end{cases}}\Rightarrow\hept{\begin{cases}x=7\\x=4\end{cases}}\)

Vậy ...

a, 2X+69.2=69.4

2X+138=276

2X=276-138=138

X=138:2=69

b,2X-12-x=0 

2X-X=12-0

x=12

mik chi biet vay thoi con cau cuoi mik chiu

2x+69.2=69.4

2x=69.4-69.2

2x=69.(4-2)

2x=69.2

x=69.2/2

x=69

2x-12-x=0

2x-x=0+12

x=12

(x-7).(2x-8)=0

ta có hai trường hợp:  

trường hợp 1:     x-7=0

                          x=7

trường hợp 2:

2x-8=0

2x=8

x=4

bạn cố gắng hiểu nha. mình làm hơi tắt.

\(a,\left(4x-1\right)\left(x^2+12\right)\left(-x+4\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1>0\\x^2+12>0\left(LD\forall x\right)\\-x+4>0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x>1\\-x>-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{4}\\x< 4\end{matrix}\right.\)

Vậy \(S=\left\{x|\dfrac{1}{4}< x< 4\right\}\)

\(b,\left(2x-1\right)\left(5-2x\right)\left(1-x\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1< 0\\5-2x< 0\\1-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{1}{2}\\x>\dfrac{5}{2}\\x< 1\end{matrix}\right.\)

Vậy \(S=\left\{x|1>x>\dfrac{5}{2}\right\}\)