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受欢迎的代数 >

a(x^2+1)=x(a^2+1)

解答

a (x^{2} + 1) = x (a^{2} + 1)

解答

x = a, x = \frac{1}{a}; a \neq = 0

求解步骤

a (x^{2} + 1) = x (a^{2} + 1)

展开

a (x^{2} + 1) : a x^{2} + a

a x^{2} + a = x (a^{2} + 1)

将

x (a^{2} + 1)

para o lado esquerdo

a x^{2} + a - x (a^{2} + 1) = 0

改写成标准形式

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

a x^{2} - (a^{2} + 1) x + a = 0

使用求根公式求解

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

化简

(- a^{2} - 1)^{2} - 4 aa : (a + 1) (a - 1)

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

将解分隔开

x_{1} = \frac{- ( - a ^{2} - 1 ) + ( a + 1 ) ( a - 1 )}{2 a}, x_{2} = \frac{- ( - a ^{2} - 1 ) - ( a + 1 ) ( a - 1 )}{2 a}

x = \frac{- ( - a ^{2} - 1 ) + ( a + 1 ) ( a - 1 )}{2 a} : a

x = \frac{- ( - a ^{2} - 1 ) - ( a + 1 ) ( a - 1 )}{2 a} : \frac{1}{a}

二次方程组的解是:

x = a, x = \frac{1}{a}; a \neq = 0

作图

流行的例子

solvefor x,f(x^2+y^2+z^2z^2-2xy)=0

so l v e f or x, f (x^{2} + y^{2} + z^{2} z^{2} - 2 x y) = 0

3 = xx + xx + xx

solvefor x,3x+4y(3x^2+4y^2)=0

so l v e f or x, 3 x + 4 y (3 x^{2} + 4 y^{2}) = 0

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((6x^2))/5 =4.8

\frac{( 6 x ^{2} )}{5} = 4.8