9x^2-5=0 x^2-x+1/4=0. [tìm x] 14x^2-7x=0 07/07/2021 Bởi Lyla 9x^2-5=0 x^2-x+1/4=0. [tìm x] 14x^2-7x=0
Đáp án: a) 9x² − 5 = 0⇒ x² = $\frac{5}{9}$ TH1: x = $\frac{\sqrt{5}}{3}$ TH2: x=-$\frac{\sqrt{5}}{3}$ b) x² − x + 1/4 = 0 ⇒ (x−$\frac{1}{2}$ )² = 0 ⇒ x=$\frac{1}{2}$ c) 14x² − 7x = 0 ⇒ 7( 2x²-x ) = 0 ⇒ 2x² − x = 0 ⇒ x( 2x−1 )=0 TH1: x=0 TH2: 2x-1=0 ⇒ x = $\frac{1}{2}$ Bình luận
$9x^2-5=0$`⇒x^2=5/9` `⇒x=(\sqrt5)/(±3)` `x^2-x+1/4=0` `⇒(x-1/2)^2=0` `⇒x=1/2` `14x^2-7x=0` `⇒2x^2-x=0` `⇒x(2x-1)=0` $⇒\left[ \begin{array}{l}x=0\\2x-1=0\end{array} \right.$ $⇒\left[ \begin{array}{l}x=0\\x=\frac{1}{2}\end{array} \right.$ Bình luận
Đáp án:
a) 9x² − 5 = 0
⇒ x² = $\frac{5}{9}$
TH1: x = $\frac{\sqrt{5}}{3}$
TH2: x=-$\frac{\sqrt{5}}{3}$
b) x² − x + 1/4 = 0
⇒ (x−$\frac{1}{2}$ )² = 0
⇒ x=$\frac{1}{2}$
c) 14x² − 7x = 0
⇒ 7( 2x²-x ) = 0
⇒ 2x² − x = 0
⇒ x( 2x−1 )=0
TH1: x=0
TH2: 2x-1=0
⇒ x = $\frac{1}{2}$
$9x^2-5=0$
`⇒x^2=5/9`
`⇒x=(\sqrt5)/(±3)`
`x^2-x+1/4=0`
`⇒(x-1/2)^2=0`
`⇒x=1/2`
`14x^2-7x=0`
`⇒2x^2-x=0`
`⇒x(2x-1)=0`
$⇒\left[ \begin{array}{l}x=0\\2x-1=0\end{array} \right.$
$⇒\left[ \begin{array}{l}x=0\\x=\frac{1}{2}\end{array} \right.$