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solvefor z,ln(x^2+y^2)+3^{xy}+z^2+z=0

解答

求解

z, ln (x^{2} + y^{2}) + 3^{x y} + z^{2} + z = 0

解答

z = \frac{- 1 + 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}, z = \frac{- 1 - 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}

求解步骤

ln (x^{2} + y^{2}) + 3^{x y} + z^{2} + z = 0

改写成标准形式

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

z^{2} + z + ln (x^{2} + y^{2}) + 3^{x y} = 0

使用求根公式求解

z_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1}

化简

1^{2} - 4 \cdot 1 \cdot (ln (x^{2} + y^{2}) + 3^{x y}) : 1 - 4 (ln (x^{2} + y^{2}) + 3^{x y})

z_{1, 2} = \frac{- 1 \pm 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1}

将解分隔开

z_{1} = \frac{- 1 + 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1}, z_{2} = \frac{- 1 - 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1}

z = \frac{- 1 + 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1} : \frac{- 1 + 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}

z = \frac{- 1 - 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2 \cdot 1} : \frac{- 1 - 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}

二次方程组的解是:

z = \frac{- 1 + 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}, z = \frac{- 1 - 1 - 4 ( ln ( x ^{2} + y ^{2} ) + 3 ^{x y} )}{2}

作图

流行的例子

10*50= 1/2*x^2+10*30

10 \cdot 50 = \frac{1}{2} \cdot x^{2} + 10 \cdot 30

4(c+7)4(c+7)=36

4 (c + 7) 4 (c + 7) = 36

8 x^{2} = - 7 n + 8

solvefor p,(y-x)(xp-y)^2=x^3

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

0 = x^{2} + 9 x + 7