`4x^2-1=0`

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

`<=>[(2x-1=0),(2x+1=0):}`

`<=>[(2x=1),(2x=-1):}`

`<=>[(x=1/2),(x=-1/2):}`

Vậy `x=1/2` hoặc `x=-1/2`

Trong đề đâu có x đâu bạn

Ta có: \(4x^2-1=0\)

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

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

(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 \(x=\pm3\)

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`.

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

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

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

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

\(9x+12=0\)

\(9x=-12\)

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

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

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

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

\(\Leftrightarrow9x+14=2\)

\(\Leftrightarrow9x=2-14\)

\(\Leftrightarrow9x=-12\)

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

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

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 : \(S=\left\{-1;2+\sqrt{3};2-\sqrt{3}\right\}\)

(x + 1)(x + 2)(x + 5) − x²(x + 8) = 27

x² + 2x + x + 2(x + 5) − x³ − 8x² = 27

x²(x + 5) + 2x(x + 5) + x(x + 5) + 2(x + 5) − x³ − 8x² = 27

x³ + 5x² + 2x² + 10x + x² + 5x + 2x + 10 − x³ − 8x² = 27

17x + 10 = 27

17x = 17

    x = 17 : 17

    x = 1

Vậy x = 1