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受欢迎的代数 >

solvefor x,f(x+1)=-x^2+x+1

解答

求解

x, f (x + 1) = - x^{2} + x + 1

解答

x = - \frac{- 1 + f + f ^{2} - 6 f + 5}{2}, x = - \frac{- 1 + f - f ^{2} - 6 f + 5}{2}

求解步骤

f (x + 1) = - x^{2} + x + 1

展开

f (x + 1) : x f + f

x f + f = - x^{2} + x + 1

交换两边

- x^{2} + x + 1 = x f + f

将

f

para o lado esquerdo

- x^{2} + x + 1 - f = x f

将

x f

para o lado esquerdo

- x^{2} + x + 1 - f - x f = 0

改写成标准形式

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

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

使用求根公式求解

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

化简

(1 - f)^{2} - 4 (- 1) (1 - f) : f^{2} - 6 f + 5

x_{1, 2} = \frac{- ( 1 - f ) \pm f ^{2} - 6 f + 5}{2 ( - 1 )}

将解分隔开

x_{1} = \frac{- ( 1 - f ) + f ^{2} - 6 f + 5}{2 ( - 1 )}, x_{2} = \frac{- ( 1 - f ) - f ^{2} - 6 f + 5}{2 ( - 1 )}

x = \frac{- ( 1 - f ) + f ^{2} - 6 f + 5}{2 ( - 1 )} : - \frac{- 1 + f + f ^{2} - 6 f + 5}{2}

x = \frac{- ( 1 - f ) - f ^{2} - 6 f + 5}{2 ( - 1 )} : - \frac{- 1 + f - f ^{2} - 6 f + 5}{2}

二次方程组的解是:

x = - \frac{- 1 + f + f ^{2} - 6 f + 5}{2}, x = - \frac{- 1 + f - f ^{2} - 6 f + 5}{2}

作图

流行的例子

solvefor x,x^2-3x^2y=C

so l v e f or x, x^{2} - 3 x^{2} y = C

3 k^{2} - 96 k + 92 = 0

(2x+10)(3x-15)=0

(2 x + 10) (3 x - 15) = 0

12 x^{2} - 2 = 8 x

(81+k(k+1))/2 =127

\frac{81 + k ( k + 1 )}{2} = 127