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受欢迎的代数 >

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

解答

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

解答

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

求解步骤

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

展开

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

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

将

a

para o lado esquerdo

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

改写成标准形式

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

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

使用求根公式求解

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}}

化简

(\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}}

将解分隔开

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}

二次方程组的解是:

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

作图

流行的例子

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

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

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

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