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受欢迎的代数 >

solvefor x,f=(m-2)x^2+(n-1)x+(p-3)

解答

求解

x, f = (m - 2) x^{2} + (n - 1) x + (p - 3)

解答

x = \frac{- n + 1 + 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}, x = \frac{- n + 1 - 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}; m \neq = 2

求解步骤

f = (m - 2) x^{2} + (n - 1) x + (p - 3)

展开

(m - 2) x^{2} + (n - 1) x + (p - 3) : m x^{2} - 2 x^{2} + n x - x + p - 3

f = m x^{2} - 2 x^{2} + n x - x + p - 3

交换两边

m x^{2} - 2 x^{2} + n x - x + p - 3 = f

将

f

para o lado esquerdo

m x^{2} - 2 x^{2} + n x - x + p - 3 - f = 0

改写成标准形式

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

(m - 2) x^{2} + (n - 1) x + p - 3 - f = 0

使用求根公式求解

x_{1, 2} = \frac{- ( n - 1 ) \pm ( n - 1 ) ^{2} - 4 ( m - 2 ) ( p - 3 - f )}{2 ( m - 2 )}

化简

(n - 1)^{2} - 4 (m - 2) (p - 3 - f) : 4 m f - 8 f + 12 m + n^{2} + 8 p - 4 m p - 2 n - 23

x_{1, 2} = \frac{- ( n - 1 ) \pm 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}; m \neq = 2

将解分隔开

x_{1} = \frac{- ( n - 1 ) + 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}, x_{2} = \frac{- ( n - 1 ) - 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}

x = \frac{- ( n - 1 ) + 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )} : \frac{- n + 1 + 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}

x = \frac{- ( n - 1 ) - 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )} : \frac{- n + 1 - 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}

二次方程组的解是:

x = \frac{- n + 1 + 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}, x = \frac{- n + 1 - 4 m f - 8 f + 12 m + n ^{2} + 8 p - 4 m p - 2 n - 23}{2 ( m - 2 )}; m \neq = 2

作图

流行的例子

x^{2} + 10 x - 375 = 0

4 w^{2} + 8 w + 4 = 0

y^{2} - 13 y + 40 = 0

x^{2} + 2 (x + 1) = 5

q^2=40000000*10

q^{2} = 40000000 \cdot 10