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受欢迎的代数 >

(3.210^{-5})(2-x)^2=(6.910^{-6})x^2

解答

(3.2 \cdot 1 0^{- 5}) (2 - x)^{2} = (6.9 \cdot 1 0^{- 6}) x^{2}

解答

x = \frac{50 ( 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}, x = \frac{50 ( - 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}

+1

十进制

x = 3.73381 \dots, x = 1.36578 \dots

求解步骤

(3.2 \cdot 1 0^{- 5}) (2 - x)^{2} = (6.9 \cdot 1 0^{- 6}) x^{2}

将

(6.9 \cdot 1 0^{- 6}) x^{2}

para o lado esquerdo

2^{2} \cdot \frac{1}{2 \cdot 5 ^{6}} - \frac{12.8 x}{1 0 ^{5}} + \frac{25.1 x ^{2}}{1000000} = 0

改写成标准形式

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

\frac{25.1 x ^{2}}{1000000} - \frac{12.8 x}{1 0 ^{5}} + \frac{2}{5 ^{6}} = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) \pm ( - \frac{12.8}{1 0 ^{5}} ) ^{2} - 4 \cdot \frac{25.1}{1000000} \cdot \frac{2}{5 ^{6}}}{2 \cdot \frac{25.1}{1000000}}

(- \frac{12.8}{1 0 ^{5}})^{2} - 4 \cdot \frac{25.1}{1000000} \cdot \frac{2}{5 ^{6}} = \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}

x_{1, 2} = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) \pm \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}}{2 \cdot \frac{25.1}{1000000}}

将解分隔开

x_{1} = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) + \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}}{2 \cdot \frac{25.1}{1000000}}, x_{2} = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) - \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}}{2 \cdot \frac{25.1}{1000000}}

x = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) + \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}}{2 \frac{25.1}{1000000}} : \frac{50 ( 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}

x = \frac{- ( - \frac{12.8}{1 0 ^{5}} ) - \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000}}{2 \frac{25.1}{1000000}} : \frac{50 ( - 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}

二次方程组的解是:

x = \frac{50 ( 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}, x = \frac{50 ( - 100000 \frac{163.84}{1 0 ^{10}} - \frac{200.8}{15625000000} + 12.8 )}{251}

作图

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