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受欢迎的代数 >

solvefor x,z(x+y)p+z(x-y)q=x^2+y^2

解答

求解

x, z (x + y) p + z (x - y) q = x^{2} + y^{2}

解答

x = \frac{z q + p z + ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}, x = \frac{z q + p z - ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}

求解步骤

z (x + y) p + z (x - y) q = x^{2} + y^{2}

展开

z (x + y) p + z (x - y) q : p x z + p zy + x z q - zy q

p x z + p zy + x z q - zy q = x^{2} + y^{2}

交换两边

x^{2} + y^{2} = p x z + p zy + x z q - zy q

将

zy q

para o lado esquerdo

x^{2} + y^{2} + zy q = p x z + p zy + x z q

将

x z q

para o lado esquerdo

x^{2} + y^{2} + zy q - x z q = p x z + p zy

两边减去

p x z + p zy

x^{2} + y^{2} + zy q - x z q - (p x z + p zy) = p x z + p zy - (p x z + p zy)

化简

x^{2} - (z q + p z) x + y^{2} + zy q - p zy = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - z q - p z ) \pm ( - z q - p z ) ^{2} - 4 \cdot 1 \cdot ( y ^{2} + zy q - p zy )}{2 \cdot 1}

化简

(- z q - p z)^{2} - 4 \cdot 1 \cdot (y^{2} + zy q - p zy) : (- z q - p z)^{2} - 4 (y^{2} + zy q - p zy)

x_{1, 2} = \frac{- ( - z q - p z ) \pm ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - z q - p z ) + ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2 \cdot 1}, x_{2} = \frac{- ( - z q - p z ) - ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2 \cdot 1}

x = \frac{- ( - z q - p z ) + ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2 \cdot 1} : \frac{z q + p z + ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}

x = \frac{- ( - z q - p z ) - ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2 \cdot 1} : \frac{z q + p z - ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}

二次方程组的解是:

x = \frac{z q + p z + ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}, x = \frac{z q + p z - ( - z q - p z ) ^{2} - 4 ( y ^{2} + zy q - p zy )}{2}

作图

流行的例子

(1/2)a(a-18)=2480

(\frac{1}{2}) a (a - 18) = 2480

n(t)=15t^2-180t+5000

n (t) = 15 t^{2} - 180 t + 5000

6x^2+(a-12)x-(a^2+a-6)=0

6 x^{2} + (a - 12) x - (a^{2} + a - 6) = 0

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

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

solvefor x,p^2+q^2=x^2+y^2

so l v e f or x, p^{2} + q^{2} = x^{2} + y^{2}