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受欢迎的代数 >

solvefor c,mba=nkc^2sp+mca

解答

求解

c, m \cdot b \cdot a = n \cdot k \cdot c^{2} \cdot s \cdot p + m \cdot c \cdot a

解答

c = \frac{- ma + m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}, c = \frac{- ma - m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}; nk s p \neq = 0

求解步骤

mba = nk c^{2} s p + m c a

交换两边

nk c^{2} s p + m c a = mba

将

m \cdot b \cdot a

para o lado esquerdo

nk c^{2} s p + m c a - mba = 0

改写成标准形式

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

nk s p c^{2} + ma c - mba = 0

使用求根公式求解

c_{1, 2} = \frac{- ma \pm ( ma ) ^{2} - 4 nk s p ( - mba )}{2 nk s p}

化简

(ma)^{2} - 4 nk s p (- mba) : m^{2} a^{2} + 4 mnkba s p

c_{1, 2} = \frac{- ma \pm m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}; nk s p \neq = 0

将解分隔开

c_{1} = \frac{- ma + m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}, c_{2} = \frac{- ma - m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}

c = \frac{- ma + m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}; nk s p \neq = 0

c = \frac{- ma - m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}; nk s p \neq = 0

二次方程组的解是:

c = \frac{- ma + m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}, c = \frac{- ma - m ^{2} a ^{2} + 4 mnkba s p}{2 nk s p}; nk s p \neq = 0

作图

流行的例子

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x^{2} + 14 x + 65 = - 2

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

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

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