`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 = {3;-3}

b. x thuộc rỗng

c. x²-4=0

    x²  = 4

    x={2;-2}

d. x²+1=82

    x²   =83

 x thuộc rỗng

e. (2x)²=6

    x thuộc rỗng

f. (x-1)²=9

  TH1: x-1=3=>x=4

  TH2: x-1=-3=>x=-2

Vậy x={4;-2}

g.(2x+3)²=25

  TH1: 2x+3=5=> x=1

  Th2: 2x+3=-5=>x=-4

VẬY X={1;-4}

a, x^2= 9 

=>\(\sqrt{9}=3\)

b,\(x^2=5=>x=\sqrt{5}\)

c, x^2-4=0

=>x^2=4

=>x=2

d, x^2+1=82

=>x^2=81 =>\(\sqrt{81}=9\)

3, 2x^2=6

=>x= \(\sqrt{6}\)

f, {x-1} ^2=9

=> x-1=3

=>x=2

g{ 2x+3}^2=25

=> 2x+3=5

=>2x=2

=>x=1

a. 3.(x-2)+2.(x-3)=13

x=5

b. (x+1).(2-x)-(3x+5).(x+2)=-4x²+1

x=-9/10

c.x.(5-2x)+2x.(x-1)=13

x=13/3

d. (2x+3)²-(x-1)²=0

x=-2/3

e. x².(3x-2)-8+12=0

x vô ngiệm

f x²+x=0

x=-1

g. x³-5x=0

x=0

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~ 

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

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

\(3x-6+2x-6=13\)

\(5x=13+6+6\)

\(5x=25\)

\(x=25\)

c)  \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(5x-2x^2+2x^2-2x=13\)

\(3x=13\)

\(x=\frac{13}{3}\)

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

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

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

\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)

f)  \(x^2+x=0\)

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

\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

g)   \(x^3-5x=0\)

\(x^2\left(x-5\right)=0\)

\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)

\(\)

a: Ta có: \(x^2-2x=0\)

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

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

b: Ta có: \(\left(x-1\right)\cdot x-2\left(1-x\right)=0\)

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

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

c: Ta có: \(x^3+2x^2+x=0\)

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

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

d: Ta có: \(x^3-3x^2=0\)

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

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

mình cần gấp

a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)

\(\Leftrightarrow2x^2-2x-2x^2=-6\)

\(\Leftrightarrow x=3\)

b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

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

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^