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受欢迎的代数 >

De=(5p-x)(5p-x)/p-2((5*p-x))/p

解答

De = (5 \cdot p - x) \cdot \frac{5 \cdot p - x}{p} - 2 \cdot \frac{( 5 \cdot p - x )}{p}

解答

x = 5 p - 1 + eD p + 1, x = 5 p - 1 - eD p + 1

求解步骤

De = (5 p - x) \frac{5 p - x}{p} - 2 \cdot \frac{( 5 p - x )}{p}

展开

(5 p - x) \frac{5 p - x}{p} - 2 \cdot \frac{5 p - x}{p} : 25 p - 10 x + \frac{x ^{2}}{p} - 10 + \frac{2 x}{p}

De = 25 p - 10 x + \frac{x ^{2}}{p} - 10 + \frac{2 x}{p}

交换两边

25 p - 10 x + \frac{x ^{2}}{p} - 10 + \frac{2 x}{p} = De

将

De

para o lado esquerdo

25 p - 10 x + \frac{x ^{2} + 2 x}{p} - 10 - eD = 0

改写成标准形式

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

\frac{1 \cdot x ^{2}}{p} + (- 10 + \frac{2}{p}) x + 25 p - 10 - eD = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 10 + \frac{2}{p} ) \pm ( - 10 + \frac{2}{p} ) ^{2} - 4 \cdot \frac{1}{p} ( 25 p - 10 - eD )}{2 \cdot \frac{1}{p}}

化简

(- 10 + \frac{2}{p})^{2} - 4 \cdot \frac{1}{p} (25 p - 10 - eD) : \frac{2 eD p + 1}{p}

x_{1, 2} = \frac{- ( - 10 + \frac{2}{p} ) \pm \frac{2 eD p + 1}{p}}{2 \cdot \frac{1}{p}}

将解分隔开

x_{1} = \frac{- ( - 10 + \frac{2}{p} ) + \frac{2 eD p + 1}{p}}{2 \cdot \frac{1}{p}}, x_{2} = \frac{- ( - 10 + \frac{2}{p} ) - \frac{2 eD p + 1}{p}}{2 \cdot \frac{1}{p}}

x = \frac{- ( - 10 + \frac{2}{p} ) + \frac{2 eD p + 1}{p}}{2 \frac{1}{p}} : 5 p - 1 + eD p + 1

x = \frac{- ( - 10 + \frac{2}{p} ) - \frac{2 eD p + 1}{p}}{2 \frac{1}{p}} : 5 p - 1 - eD p + 1

二次方程组的解是:

x = 5 p - 1 + eD p + 1, x = 5 p - 1 - eD p + 1

作图

流行的例子

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