Giải phương trình (x + 2)(3 – 4x) = x^2 + 4x + 4 02/08/2021 Bởi Adalyn Giải phương trình (x + 2)(3 – 4x) = x^2 + 4x + 4
`(x + 2)(3 – 4x) = x^2 + 4x + 4 ` `⇒ (x+2)(3-4x)=(x+2)²` `⇒ (x+2)(3-4x)-(x+2)²=0` `⇒ (x+2)(3-4x-x-2)=0` `⇒ (x+2)(-5x+1)=0` `⇒ x+2=0⇒x=-2` `5x+1=0⇒4x=-1⇒x=-1/5` Bình luận
`(x + 2)(3 – 4x) = (x + 2)^2 ` `(x + 2)(3 – 4x) – (x + 2)^2 = 0` `(x + 2)(3 – 4x – x – 2) = 0` `(x + 2)( – 5x + 1) = 0` ⇔\(\left[ \begin{array}{l}x+2=0\\– 5x + 1=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=-2\\x=\frac{1}{5}\end{array} \right.\) Bình luận
`(x + 2)(3 – 4x) = x^2 + 4x + 4 `
`⇒ (x+2)(3-4x)=(x+2)²`
`⇒ (x+2)(3-4x)-(x+2)²=0`
`⇒ (x+2)(3-4x-x-2)=0`
`⇒ (x+2)(-5x+1)=0`
`⇒ x+2=0⇒x=-2`
`5x+1=0⇒4x=-1⇒x=-1/5`
`(x + 2)(3 – 4x) = (x + 2)^2 `
`(x + 2)(3 – 4x) – (x + 2)^2 = 0`
`(x + 2)(3 – 4x – x – 2) = 0`
`(x + 2)( – 5x + 1) = 0`
⇔\(\left[ \begin{array}{l}x+2=0\\– 5x + 1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-2\\x=\frac{1}{5}\end{array} \right.\)