\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

x²+4x+4=0
(x+2)²=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)²+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0           Th2: x+1=0
            x=-3                      x=-1
vậy ...

\(x-\frac{2}{8}=\frac{-1}{4}\)

\(\Leftrightarrow x=\frac{-1}{4}+\frac{2}{8}\)

\(\Leftrightarrow x=\frac{-1}{4}+\frac{1}{4}\)

\(\Leftrightarrow x=0\)

Vậy ....

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

=>3(x-2)=4(x+5)

=>4x+20=3x-6

=>x=-26

`(x-2):4 = (5+x) :3`

`(x-2)*3 = 4*(5+x)`

`3x -6 = 20 +4x`

`3x -4x = 20 +6`

`-x =26`

`x=26`

=> 2x-2 + 3x-6 = x-4

=> 5x-8 = x-4

=> 5x-8-(x-4) = 0

=> 5x-8-x+4=0

=> 4x-4=0

=> 4x=4

=> x=4:4=1

Vậy x=1

Tk mk nha

2(x-1)+3(x-2)=x-4

<=>2x-2+3x-6=x-4

<=>5x-8=x-4

<=>5x-x=8-4

<=>4x=4

<=>x=1

a

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

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

(x+3)(x-2)+3x=4(x+3/4)

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

x²+3x-2x-6+3x-4x=3

x²+(3x-2x+3x-4x)-6=3

x²-6=3

x²=9

x²=3²

=>x=|3|

x=3 hoặc x=-3

\(\left(x+3\right)\left(x-2\right)+3x=4\left(x+\frac{3}{4}\right)\)

\(\Rightarrow x^2-2x+3x-6+3x=4x+3\)

\(\Rightarrow x^2+4x-6=4x+3\)

\(\Rightarrow x^2-6=3\)

\(\Rightarrow x^2=3+6=9\)

\(\Rightarrow x=-3\)hoặc \(x=3\)