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 >

solvefor x,f(x^2+1)=5x+3

解

解く

x, f (x^{2} + 1) = 5 x + 3

解

x = \frac{5 + 25 - 4 f ( f - 3 )}{2 f}, x = \frac{5 - 25 - 4 f ( f - 3 )}{2 f}; f \neq = 0

解答ステップ

f (x^{2} + 1) = 5 x + 3

拡張

f (x^{2} + 1) : x^{2} f + f

x^{2} f + f = 5 x + 3

3

を左側に移動します

x^{2} f + f - 3 = 5 x

5 x

を左側に移動します

x^{2} f + f - 3 - 5 x = 0

標準的な形式で書く

a x^{2} + b x + c = 0

f x^{2} - 5 x + f - 3 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 f ( f - 3 )}{2 f}

簡素化

(- 5)^{2} - 4 f (f - 3) : 25 - 4 f (f - 3)

x_{1, 2} = \frac{- ( - 5 ) \pm 25 - 4 f ( f - 3 )}{2 f}; f \neq = 0

解を分離する

x_{1} = \frac{- ( - 5 ) + 25 - 4 f ( f - 3 )}{2 f}, x_{2} = \frac{- ( - 5 ) - 25 - 4 f ( f - 3 )}{2 f}

x = \frac{- ( - 5 ) + 25 - 4 f ( f - 3 )}{2 f} : \frac{5 + 25 - 4 f ( f - 3 )}{2 f}

x = \frac{- ( - 5 ) - 25 - 4 f ( f - 3 )}{2 f} : \frac{5 - 25 - 4 f ( f - 3 )}{2 f}

二次equationの解:

x = \frac{5 + 25 - 4 f ( f - 3 )}{2 f}, x = \frac{5 - 25 - 4 f ( f - 3 )}{2 f}; f \neq = 0

グラフ

人気の例

\frac{x ^{2}}{2} + 2 x + 2 = 0

(-1)/(500)(x-100)^2+20=0

\frac{- 1}{500} (x - 100)^{2} + 20 = 0

5 x^{2} + x - \frac{1}{5} = 0

v^{2} = 32

30 - (4 x + 3)^{2} = 48