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人気のある代数 >

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

解

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

解

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

解答ステップ

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

拡張

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

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

x (a^{2} + 1)

を左側に移動します

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

標準的な形式で書く

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

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

解くとthe二次式

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

簡素化

(- 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

解を分離する

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equationの解:

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

グラフ

人気の例

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

4x^2-2(a^4-b^4)x+a^2b^2=0

4 x^{2} - 2 (a^{4} - b^{4}) x + a^{2} b^{2} = 0

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

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