abcx^2-(a^2b^2+c^2)x+abc=0

解

ab c x^{2} - (a^{2} b^{2} + c^{2}) x + ab c = 0

x = \frac{a ^{2} b ^{2} + c ^{2} + a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}, x = \frac{a ^{2} b ^{2} + c ^{2} - a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}; ab c \neq = 0

解答ステップ

ab c x^{2} - (a^{2} b^{2} + c^{2}) x + ab c = 0

解くとthe二次式

x_{1, 2} = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) \pm ( - a ^{2} b ^{2} - c ^{2} ) ^{2} - 4 ab c ab c}{2 ab c}

簡素化

(- a^{2} b^{2} - c^{2})^{2} - 4 ab c ab c : a^{4} b^{4} - 2 a^{2} b^{2} c^{2} + c^{4}

x_{1, 2} = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) \pm a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}; ab c \neq = 0

解を分離する

x_{1} = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) + a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}, x_{2} = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) - a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}

x = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) + a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c} : \frac{a ^{2} b ^{2} + c ^{2} + a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}

x = \frac{- ( - a ^{2} b ^{2} - c ^{2} ) - a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c} : \frac{a ^{2} b ^{2} + c ^{2} - a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}

二次equationの解:

x = \frac{a ^{2} b ^{2} + c ^{2} + a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}, x = \frac{a ^{2} b ^{2} + c ^{2} - a ^{4} b ^{4} - 2 a ^{2} b ^{2} c ^{2} + c ^{4}}{2 ab c}; ab c \neq = 0