求解 {l}{x^2+2y^2=8}{y=m(x-2)+sqrt{2}}

$$\left. \begin{array} { l } { x ^ { 2 } + 2 y ^ { 2 } = 8 } \\ { y = m ( x - 2 ) + \sqrt { 2 } } \end{array} \right.$$

求解 x, y 的值

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(2m|\frac{\sqrt{2}\left(\sqrt{2}m+1\right)}{2}|-\sqrt{2}m+1\right)}{2m^{2}+1}$$

运用代入法和二次公式的步骤

考虑第 2 个公式。 使用分配律将 m 乘以 x-2。

$$y=mx-2m+\sqrt{2}$$

用 mx-2m+\sqrt{2} 替换另一个方程式中 x^{2}+2y^{2}=8 中的 y。

$$x^{2}+2\left(mx-2m+\sqrt{2}\right)^{2}=8$$

对 mx-2m+\sqrt{2} 进行平方运算。

$$x^{2}+2\left(m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2}\right)=8$$

求 2 与 m^{2}x^{2}+2m\left(-2m+\sqrt{2}\right)x+\left(-2m+\sqrt{2}\right)^{2} 的乘积。

$$x^{2}+2m^{2}x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

将 2m^{2}x^{2} 加上 x^{2}。

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}=8$$

将等式的两边同时减去 8。

$$\left(2m^{2}+1\right)x^{2}+4m\left(-2m+\sqrt{2}\right)x+2\left(-2m+\sqrt{2}\right)^{2}-8=0$$

此公式采用标准形式: ax^{2}+bx+c=0。在二次公式 \frac{-b±\sqrt{b^{2}-4ac}}{2a} 中用 1+2m^{2} 替换 a，2\times 2m\left(-2m+\sqrt{2}\right) 替换 b，并用 -4+8m^{2}-8m\sqrt{2} 替换 c。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{\left(4m\left(-2m+\sqrt{2}\right)\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

对 2\times 2m\left(-2m+\sqrt{2}\right) 进行平方运算。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-4\left(2m^{2}+1\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

求 -4 与 1+2m^{2} 的乘积。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}+\left(-8m^{2}-4\right)\left(8m^{2}-8\sqrt{2}m-4\right)}}{2\left(2m^{2}+1\right)}$$

求 -4-8m^{2} 与 -4+8m^{2}-8m\sqrt{2} 的乘积。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{16m^{2}\left(-2m+\sqrt{2}\right)^{2}-64m^{4}+64\sqrt{2}m^{3}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

将 16+32m\sqrt{2}-64m^{4}+64m^{3}\sqrt{2} 加上 16m^{2}\left(-2m+\sqrt{2}\right)^{2}。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±\sqrt{32m^{2}+32\sqrt{2}m+16}}{2\left(2m^{2}+1\right)}$$

取 16+32m^{2}+32m\sqrt{2} 的平方根。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{2\left(2m^{2}+1\right)}$$

求 2 与 1+2m^{2} 的乘积。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

现在 ± 为加号时求公式 x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} 的解。 将 4\sqrt{1+2m^{2}+2m\sqrt{2}} 加上 -4m\left(-2m+\sqrt{2}\right)。

$$x=\frac{-4m\left(-2m+\sqrt{2}\right)+4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2}$$

-4m\left(-2m+\sqrt{2}\right)+4\sqrt{1+2m^{2}+2m\sqrt{2}} 除以 2+4m^{2}。

$$x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

现在 ± 为减号时求公式 x=\frac{-4m\left(-2m+\sqrt{2}\right)±4\sqrt{2m^{2}+2\sqrt{2}m+1}}{4m^{2}+2} 的解。 将 -4m\left(-2m+\sqrt{2}\right) 减去 4\sqrt{1+2m^{2}+2m\sqrt{2}}。

$$x=\frac{8m^{2}-4\sqrt{2m^{2}+2\sqrt{2}m+1}-4\sqrt{2}m}{4m^{2}+2}$$

8m^{2}-4m\sqrt{2}-4\sqrt{1+2m^{2}+2m\sqrt{2}} 除以 2+4m^{2}。

$$x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

x 有两个解: \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} 和 \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}}。用 \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} 替换等式 y=mx-2m+\sqrt{2} 中的 x，可求得同时满足两个方程式的 y 的相应解。

$$y=m\times \frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

求 m 与 \frac{2\left(2m^{2}-m\sqrt{2}+\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} 的乘积。

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

现在用 \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} 替换等式 y=mx-2m+\sqrt{2} 中的 x，并求得可同时满足两个等式的 y 的相应解。

$$y=m\times \frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}-2m+\sqrt{2}$$

求 m 与 \frac{2\left(2m^{2}-m\sqrt{2}-\sqrt{2m^{2}+1+2m\sqrt{2}}\right)}{1+2m^{2}} 的乘积。

$$y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2}$$

系统现在已得到解决。

$$y=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{ or }y=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}m-2m+\sqrt{2},x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}$$

求解 x, y 的值 (复数求解)

$$\left\{\begin{matrix}x=\frac{2\left(2m^{2}-\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(-m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{; }x=\frac{2\left(2m^{2}+\sqrt{2m^{2}+2\sqrt{2}m+1}-\sqrt{2}m\right)}{2m^{2}+1}\text{, }y=\frac{\sqrt{2}\left(m\sqrt{2\left(\sqrt{2}m+1\right)^{2}}-\sqrt{2}m+1\right)}{2m^{2}+1}\text{, }&m\neq -\frac{\sqrt{2}i}{2}\text{ and }m\neq \frac{\sqrt{2}i}{2}\\x=-\frac{\left(-2m+\sqrt{2}\right)^{2}-4}{2m\left(-2m+\sqrt{2}\right)}\text{, }y=\frac{2m^{2}-2\sqrt{2}m+3}{-2m+\sqrt{2}}\text{, }&m=-\frac{\sqrt{2}i}{2}\text{ or }m=\frac{\sqrt{2}i}{2}\end{matrix}\right.$$