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人気のある代数 >

solvefor x,(C-x)(1+kx)=B(kqx)

解

解く

x, (C - x) (1 + k x) = B (k q x)

解

x = - \frac{- k C + 1 + k Bq + k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}, x = - \frac{- k C + 1 + k Bq - k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}; k \neq = 0

解答ステップ

(C - x) (1 + k x) = B (k q x)

拡張

(C - x) (1 + k x) : C + k x C - x - k x^{2}

C + k x C - x - k x^{2} = k x Bq

k x Bq

を左側に移動します

C + k x C - x - k x^{2} - k x Bq = 0

標準的な形式で書く

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

- k x^{2} + (k C - 1 - k Bq) x + C = 0

解くとthe二次式

x_{1, 2} = \frac{- ( k C - 1 - k Bq ) \pm ( k C - 1 - k Bq ) ^{2} - 4 ( - k ) C}{2 ( - k )}

簡素化

(k C - 1 - k Bq)^{2} - 4 (- k) C : k^{2} C^{2} + 2 k C - 2 k^{2} CBq + k^{2} B^{2} q^{2} + 2 k Bq + 1

x_{1, 2} = \frac{- ( k C - 1 - k Bq ) \pm k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 ( - k )}; k \neq = 0

解を分離する

x_{1} = \frac{- ( k C - 1 - k Bq ) + k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 ( - k )}, x_{2} = \frac{- ( k C - 1 - k Bq ) - k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 ( - k )}

x = \frac{- ( k C - 1 - k Bq ) + k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 ( - k )} : - \frac{- k C + 1 + k Bq + k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}

x = \frac{- ( k C - 1 - k Bq ) - k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 ( - k )} : - \frac{- k C + 1 + k Bq - k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}

二次equationの解:

x = - \frac{- k C + 1 + k Bq + k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}, x = - \frac{- k C + 1 + k Bq - k ^{2} C ^{2} + 2 k C - 2 k ^{2} CBq + k ^{2} B ^{2} q ^{2} + 2 k Bq + 1}{2 k}; k \neq = 0

グラフ

人気の例

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