ແກ້ (V_{1}-V_{2})x(V_{1}+v_{2})=2v_{1}xv_{2}

$$( V _ { 1 } - V _ { 2 } ) x ( V _ { 1 } + v _ { 2 } ) = 2 v _ { 1 } x v _ { 2 }$$

ແກ້ສຳລັບ V_2 (complex solution)

$$\left\{\begin{matrix}V_{2}=\frac{V_{1}^{2}+V_{1}v_{2}-2v_{1}v_{2}}{v_{2}+V_{1}}\text{, }&V_{1}\neq -v_{2}\\V_{2}\in \mathrm{C}\text{, }&\left(v_{1}=0\text{ and }V_{1}=-v_{2}\right)\text{ or }\left(v_{2}=0\text{ and }V_{1}=0\right)\text{ or }x=0\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ V_{1}-V_{2} ດ້ວຍ x.

$$\left(V_{1}x-V_{2}x\right)\left(V_{1}+v_{2}\right)=2v_{1}xv_{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ V_{1}x-V_{2}x ດ້ວຍ V_{1}+v_{2}.

$$xV_{1}^{2}+V_{1}xv_{2}-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}$$

ລົບ xV_{1}^{2} ອອກຈາກທັງສອງຂ້າງ.

$$V_{1}xv_{2}-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}-xV_{1}^{2}$$

ລົບ V_{1}xv_{2} ອອກຈາກທັງສອງຂ້າງ.

$$-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}-xV_{1}^{2}-V_{1}xv_{2}$$

ຈັດລຳດັບພົດຄືນໃໝ່.

$$-V_{1}V_{2}x-V_{2}v_{2}x=2v_{1}v_{2}x-V_{1}v_{2}x-xV_{1}^{2}$$

ຮວມທຸກຄຳສັບທີ່ມີ V_{2}.

$$\left(-V_{1}x-v_{2}x\right)V_{2}=2v_{1}v_{2}x-V_{1}v_{2}x-xV_{1}^{2}$$

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$\left(-V_{1}x-v_{2}x\right)V_{2}=2v_{1}v_{2}x-xV_{1}^{2}-V_{1}v_{2}x$$

ຫານທັງສອງຂ້າງດ້ວຍ -V_{1}x-v_{2}x.

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

ການຫານດ້ວຍ -V_{1}x-v_{2}x ຈະຍົກເລີກການຄູນດ້ວຍ -V_{1}x-v_{2}x.

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

ຫານ x\left(2v_{1}v_{2}-V_{1}v_{2}-V_{1}^{2}\right) ດ້ວຍ -V_{1}x-v_{2}x.

$$V_{2}=-\frac{2v_{1}v_{2}-V_{1}v_{2}-V_{1}^{2}}{v_{2}+V_{1}}$$

ແກ້ສຳລັບ V_2

$$\left\{\begin{matrix}V_{2}=\frac{V_{1}^{2}+V_{1}v_{2}-2v_{1}v_{2}}{v_{2}+V_{1}}\text{, }&V_{1}\neq -v_{2}\\V_{2}\in \mathrm{R}\text{, }&\left(v_{1}=0\text{ and }V_{1}=-v_{2}\right)\text{ or }\left(v_{2}=0\text{ and }V_{1}=0\right)\text{ or }x=0\end{matrix}\right,$$

ຂັ້ນຕອນສຳລັບການແກ້ສົມຜົນເສັ້ນຊື່

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ V_{1}-V_{2} ດ້ວຍ x.

$$\left(V_{1}x-V_{2}x\right)\left(V_{1}+v_{2}\right)=2v_{1}xv_{2}$$

ໃຊ້ຄຸນສົມບັດການແຈກແຈງເພື່ອຄູນ V_{1}x-V_{2}x ດ້ວຍ V_{1}+v_{2}.

$$xV_{1}^{2}+V_{1}xv_{2}-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}$$

ລົບ xV_{1}^{2} ອອກຈາກທັງສອງຂ້າງ.

$$V_{1}xv_{2}-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}-xV_{1}^{2}$$

ລົບ V_{1}xv_{2} ອອກຈາກທັງສອງຂ້າງ.

$$-V_{2}xV_{1}-V_{2}xv_{2}=2v_{1}xv_{2}-xV_{1}^{2}-V_{1}xv_{2}$$

ຈັດລຳດັບພົດຄືນໃໝ່.

$$-V_{1}V_{2}x-V_{2}v_{2}x=2v_{1}v_{2}x-V_{1}v_{2}x-xV_{1}^{2}$$

ຮວມທຸກຄຳສັບທີ່ມີ V_{2}.

$$\left(-V_{1}x-v_{2}x\right)V_{2}=2v_{1}v_{2}x-V_{1}v_{2}x-xV_{1}^{2}$$

ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.

$$\left(-V_{1}x-v_{2}x\right)V_{2}=2v_{1}v_{2}x-xV_{1}^{2}-V_{1}v_{2}x$$

ຫານທັງສອງຂ້າງດ້ວຍ -V_{1}x-v_{2}x.

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

ການຫານດ້ວຍ -V_{1}x-v_{2}x ຈະຍົກເລີກການຄູນດ້ວຍ -V_{1}x-v_{2}x.

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

ຫານ x\left(2v_{1}v_{2}-V_{1}v_{2}-V_{1}^{2}\right) ດ້ວຍ -V_{1}x-v_{2}x.

$$V_{2}=-\frac{2v_{1}v_{2}-V_{1}v_{2}-V_{1}^{2}}{v_{2}+V_{1}}$$

ແກ້ສຳລັບ V_1 (complex solution)

$$\left\{\begin{matrix}\\V_{1}=\frac{\sqrt{8v_{1}v_{2}+v_{2}^{2}+2V_{2}v_{2}+V_{2}^{2}}+V_{2}-v_{2}}{2}\text{; }V_{1}=\frac{-\sqrt{8v_{1}v_{2}+v_{2}^{2}+2V_{2}v_{2}+V_{2}^{2}}+V_{2}-v_{2}}{2}\text{, }&\text{unconditionally}\\V_{1}\in \mathrm{C}\text{, }&x=0\end{matrix}\right,$$

ແກ້ສຳລັບ V_1

$$\left\{\begin{matrix}V_{1}=\frac{\sqrt{8v_{1}v_{2}+v_{2}^{2}+2V_{2}v_{2}+V_{2}^{2}}+V_{2}-v_{2}}{2}\text{; }V_{1}=\frac{-\sqrt{8v_{1}v_{2}+v_{2}^{2}+2V_{2}v_{2}+V_{2}^{2}}+V_{2}-v_{2}}{2}\text{, }&\left(v_{1}>0\text{ and }v_{2}>0\right)\text{ or }\left(v_{2}<0\text{ and }v_{1}<0\right)\text{ or }V_{2}\geq 2\sqrt{-2v_{1}v_{2}}-v_{2}\text{ or }V_{2}\leq -2\sqrt{-2v_{1}v_{2}}-v_{2}\text{ or }\left(v_{2}\leq 0\text{ and }v_{1}\leq 0\right)\text{ or }\left(v_{1}\geq 0\text{ and }v_{2}\geq 0\right)\\V_{1}\in \mathrm{R}\text{, }&x=0\end{matrix}\right,$$