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人気のある代数 >

y((1+y)/(2-k))=a

解

y (\frac{1 + y}{2 - k}) = a

解

y = \frac{- 1 + 1 + 4 a ( - k + 2 )}{2}, y = \frac{- 1 - 1 + 4 a ( - k + 2 )}{2}

解答ステップ

y (\frac{1 + y}{2 - k}) = a

拡張

y (\frac{1 + y}{2 - k}) : \frac{y + y ^{2}}{2 - k}

\frac{y + y ^{2}}{2 - k} = a

a

を左側に移動します

\frac{y + y ^{2}}{2 - k} - a = 0

標準的な形式で書く

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

\frac{1 \cdot y ^{2}}{2 - k} + \frac{1 \cdot y}{2 - k} - a = 0

解くとthe二次式

y_{1, 2} = \frac{- \frac{1}{2 - k} \pm ( \frac{1}{2 - k} ) ^{2} - 4 \cdot \frac{1}{2 - k} ( - a )}{2 \cdot \frac{1}{2 - k}}

簡素化

(\frac{1}{2 - k})^{2} - 4 \cdot \frac{1}{2 - k} (- a) : \frac{1 + 4 a ( - k + 2 )}{- k + 2}

y_{1, 2} = \frac{- \frac{1}{2 - k} \pm \frac{1 + 4 a ( - k + 2 )}{- k + 2}}{2 \cdot \frac{1}{2 - k}}

解を分離する

y_{1} = \frac{- \frac{1}{2 - k} + \frac{1 + 4 a ( - k + 2 )}{- k + 2}}{2 \cdot \frac{1}{2 - k}}, y_{2} = \frac{- \frac{1}{2 - k} - \frac{1 + 4 a ( - k + 2 )}{- k + 2}}{2 \cdot \frac{1}{2 - k}}

y = \frac{- \frac{1}{2 - k} + \frac{1 + 4 a ( - k + 2 )}{- k + 2}}{2 \frac{1}{2 - k}} : \frac{- 1 + 1 + 4 a ( - k + 2 )}{2}

y = \frac{- \frac{1}{2 - k} - \frac{1 + 4 a ( - k + 2 )}{- k + 2}}{2 \frac{1}{2 - k}} : \frac{- 1 - 1 + 4 a ( - k + 2 )}{2}

二次equationの解:

y = \frac{- 1 + 1 + 4 a ( - k + 2 )}{2}, y = \frac{- 1 - 1 + 4 a ( - k + 2 )}{2}

グラフ

人気の例

((x+2)^2)/5+((x+2)^2)/3 = 16/3

\frac{( x + 2 ) ^{2}}{5} + \frac{( x + 2 ) ^{2}}{3} = \frac{16}{3}

0.000006x^2-0.02x+25=10-0.04x

0.000006 x^{2} - 0.02 x + 25 = 10 - 0.04 x

6 p^{2} - 10 = - 57

0=-20x^2+360x+1000

0 = - 20 x^{2} + 360 x + 1000

45 = \frac{1}{2} (5) v^{2}