Ta có: \(\left\{{}\begin{matrix}80x+160y=20\\2x+6y=0,7\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}4x+8y=1\left(1\right)\\4x+12y=1,4\left(2\right)\end{matrix}\right.\)

Trừ (1) cho (2) vế theo vế ta được:

\(12y-8y=1,4-1\) <=> \(4y=0,4\)

<=> \(y=0,1\) => \(x=\dfrac{0,7-6.0,1}{2}\) = 0,05

b) \(2\left(a-b\right)\left(c-b\right)+2\left(b-a\right)\left(c-a\right)+2\left(b-c\right)\left(a-c\right)\)

= \(2\left(ac-ab-bc+b^2+bc-ab-ac+a^2+ab-bc-ac+c^2\right)\)

= \(2a^2+2b^2+2c^2-2ab-2bc-2ac\)

= \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)

= \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\)

a) (a + b + c)² + a² + b² + c²

= a² + b² + c² + 2ab + 2bc + 2ac +a² + b² + c²

= (a² + 2ab + b² ) + (b² + 2bc + c² ) + (a² + 2ac + c²)

= (a + b)² + (b + c)² + (c+a)²