Mất dấu nên xét 2 th.

TH1

`x^2-4x+4=0`

`<=>x^2-2.x.2+2^2=0`

`<=>(x-2)^2=0`

`<=>x-2=0`

`<=>x=2`

`=>S={2}`

TH2

`x^2+4x+4=0`

`<=>x^2+2.x.2+2^2=0`

`<=>(x+2)^2=0`

`<=>x+2=0`

`<=>x=-2`

`=>S={-2}`

đề à -4x hay +4x nó sẽ ra kết quả khác nhau em ạ!

=x^2 +2.x.2 +2^2

=(x+2)^2

`x^2+4x+4=0`

`⇔x^2+2.x.2+2^2=0`

`⇔(x+2)^2=0`

`⇔x+2=0⇔x=−2`

Vậy `x=-2`.

\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2=0\Rightarrow x+2=0\Rightarrow x=-2\)

`(x-1)^2-25=0`

`<=>(x-1-5)(x-1+5)=0`

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

`<=>[(x-6=0),(x+4=0):}`

`<=>[(x=6),(x=-4):}`

Vậy `x=6` hoặc `x=-4`

Ta có: \(\left(x-1\right)^2-25=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

(2x+1)^2=0

2x=-1

x=-1/2

`4x^2+4x+1=0`

`<=>(2x)^2+2.2x+1=0`

`<=>(2x+1)^2=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2.`

\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2=0\Rightarrow2x+1=0\Rightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

x^2 -3^2=0

(x+3).(x-3)=0

x+3=0 hoặc x-3=0

x=-3 hoặc x=3

        \(x^2-9=0\)

⇔ \(x^2-3^2=0\)

⇔ \(\left(x-3\right)\left(x+3\right)=0\)

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

Vậy \(x=\pm3\)

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6x+12+3x-2=0\)

\(1+1+6x+3x+12-2=0\)

\(9x+12=0\)

\(9x=-12\)

\(x=\frac{-4}{3}\)

\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=0+2\)

\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=2\)

\(\Leftrightarrow9x+14=2\)

\(\Leftrightarrow9x=2-14\)

\(\Leftrightarrow9x=-12\)

\(\Leftrightarrow x=\frac{-12}{9}=\frac{-4}{3}\)

\(\Rightarrow x=\frac{-4}{2}\)

\(a\text{) }pt\Leftrightarrow\left(y^2+2y+1\right)+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)

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

\(\Leftrightarrow y+1=0\text{ và }2^x-1=0\)

\(\Leftrightarrow y=-1\text{ và }x=0\)

\(b\text{) }pt\Leftrightarrow\left(4x^2+4y^2+8xy\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow x+y=0\text{ và }x-1=0\text{ và }y+1=0\)

\(\Leftrightarrow x=1\text{ và }y=-1\)

Đề sai rồi bạn

Ta có :

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

\(\Leftrightarrow x^3+x^2-4x^2-4x+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\pm\sqrt{3}\end{cases}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-1;2+\sqrt{3};2-\sqrt{3}\right\}\)