Question

4x²+4x+1=x²

Answer 1

   `4x^2+4x+1=x^2`

`<=>3x^2+4x+1=0`

`<=>3x^2+3x+x+1=0`

`<=>3x(x+1)+(x+1)=0`

`<=>(x+1)(3x+1)=0`

`<=>` $\left[\begin{matrix} x+1=0\\ 3x+1=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=-1\\ x=\dfrac{-1}{3}\end{matrix}\right.$

Vậy `S={-1;[-1]/3}`

Answer 2

4x²+4x+1=x²

^{\(\Leftrightarrow3x^2+4x+1=0\)}

^{\(\Leftrightarrow3x^2+3x+x+1=0\)}

^{\(\Leftrightarrow3x\left(x+1\right)+\left(x+1\right)=0\)}

^{\(\Leftrightarrow\left(3x+1\right)\left(x+1\right)=0\)}

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

Answer 3

4x² + 4x + 1 = x²

4x² + 4x + 1 - x² = 0

3x² + 4x + 1 = 0

a + b = 4; ab = 3 . 1 = 3

a = 1; b = 3 

( 3x² + x ) + ( 3x + 1 )

x( 3x + 1 ) + 3x + 1

( 3x + 1 ) ( x + 1 )

x = -1/3 hoặc x = -1.

Answer 4

\(4x^2+4x+1=x^2\)

\(\Leftrightarrow\left[\left(2x\right)^2+2.2x.1+1^2\right]-x^2=0\)

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

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

\(\Leftrightarrow\left(x+1\right).\left(3x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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