ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

-3(2x-1)(2x+1)=2

解

- 3 (2 x - 1) (2 x + 1) = 2

解

x = - \frac{1}{2 3}, x = \frac{1}{2 3}

+1

十進法表記

x = - 0.28867 \dots, x = 0.28867 \dots

解答ステップ

- 3 (2 x - 1) (2 x + 1) = 2

拡張

- 3 (2 x - 1) (2 x + 1) : - 12 x^{2} + 3

- 12 x^{2} + 3 = 2

2

を左側に移動します

- 12 x^{2} + 1 = 0

解くとthe二次式

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 12 ) \cdot 1}{2 ( - 12 )}

0^{2} - 4 (- 12) \cdot 1 = 43

x_{1, 2} = \frac{- 0 \pm 4 3}{2 ( - 12 )}

解を分離する

x_{1} = \frac{- 0 + 4 3}{2 ( - 12 )}, x_{2} = \frac{- 0 - 4 3}{2 ( - 12 )}

x = \frac{- 0 + 4 3}{2 ( - 12 )} : - \frac{1}{2 3}

x = \frac{- 0 - 4 3}{2 ( - 12 )} : \frac{1}{2 3}

二次equationの解:

x = - \frac{1}{2 3}, x = \frac{1}{2 3}

グラフ

人気の例

6 - 7 x^{2} = 7

x^2-x=((x+1)(x+3))/2-2

x^{2} - x = \frac{( x + 1 ) ( x + 3 )}{2} - 2

n^{2} + 5 n + 7 = 7

solvefor x,5x^2+2y^2=0

so l v e f or x, 5 x^{2} + 2 y^{2} = 0

solvefor x,f(a+b)=x^2-4x

so l v e f or x, f (a + b) = x^{2} - 4 x