( 2x – 1 )² – ( x + 3 )² = 0 ⇔ 4x² – 4x + 1 – ( x² + 6x + 9 ) = 0 ⇔ 4x² – 4x + 1 – x² – 6x – 9 = 0 ⇔ 3x² – 10x – 8 = 0 ⇔ 3x² – 12x + 2x – 8 = 0 ⇔ 3x( x – 4 ) + 2( x – 4 ) = 0 ⇔ ( x – 4 )( 3x + 2 ) = 0 ⇔ \(\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=4\\x=\frac{-2}{3}\end{array} \right.\) ~CHÚC BẠN HỌC TỐT~ Bình luận
`( 2x – 1 )² – ( x + 3 )² = 0` `⇔ 4x² – 4x + 1 – ( x² + 6x + 9 ) = 0` `⇔ 4x² – 4x + 1 – x² – 6x – 9 = 0` `⇔ 3x² – 12x + 2x – 8 = 0` `⇔ 3x( x – 4 ) + 2( x – 4 ) = 0` `⇔ ( x – 4 )( 3x + 2 ) = 0` $⇔\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.$ $⇔\left[ \begin{array}{l}x=4\\x=\dfrac{-2}{3}\end{array} \right.$ Vậy `x=4` và `x=-2/3` Xin hay nhất ! Bình luận
( 2x – 1 )² – ( x + 3 )² = 0
⇔ 4x² – 4x + 1 – ( x² + 6x + 9 ) = 0
⇔ 4x² – 4x + 1 – x² – 6x – 9 = 0
⇔ 3x² – 10x – 8 = 0
⇔ 3x² – 12x + 2x – 8 = 0
⇔ 3x( x – 4 ) + 2( x – 4 ) = 0
⇔ ( x – 4 )( 3x + 2 ) = 0
⇔ \(\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\x=\frac{-2}{3}\end{array} \right.\)
~CHÚC BẠN HỌC TỐT~
`( 2x – 1 )² – ( x + 3 )² = 0`
`⇔ 4x² – 4x + 1 – ( x² + 6x + 9 ) = 0`
`⇔ 4x² – 4x + 1 – x² – 6x – 9 = 0`
`⇔ 3x² – 12x + 2x – 8 = 0`
`⇔ 3x( x – 4 ) + 2( x – 4 ) = 0`
`⇔ ( x – 4 )( 3x + 2 ) = 0`
$⇔\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=4\\x=\dfrac{-2}{3}\end{array} \right.$
Vậy `x=4` và `x=-2/3`
Xin hay nhất !