ఇక్కడ ఇంకో వ్యాసం రాయవచ్చు b ∂ 2 ∂ b f ( x ) f ( x ) = ( a + b ) 2 = a 2 + 2 a b + b 2 { 3 x + 5 y + z 7 x − 2 y + 4 z − 6 x + 3 y + 2 z = ( a + b ) 2 = a 2 + 2 a b + b 2 x 1 ∂ x 2 y {\textstyle b{\partial ^{2} \over \partial \operatorname {b} {\begin{aligned}f(x)&{\begin{alignedat}{2}f(x)&=(a+b)^{2}\\&=a^{2}+2ab+b^{2}\\\end{alignedat}}{\begin{cases}3x+5y+z\\7x-2y+4z\\-6x+3y+2z\end{cases}}=(a+b)^{2}\\&=a^{2}+2ab+b^{2}\\\end{aligned}}x_{1}\partial x_{2}}y} CO 2 + C ⟶ 2 CO {\displaystyle {\ce {CO2 + C -> 2 CO}}} 1 3 Ω 3 5 {\displaystyle {}_{1}^{3}\!\Omega _{3}^{5}} (x, y ∈ A ∪ B; x ≠ y) x² − y² ≥ 0
