Bài 1:
\(\left(2x+3\right)^2-\left(2x+3\right)\left(4x-6\right)+\left(2x-3\right)^2+xy\)
\(=\left(2x+3\right)^2-2\cdot\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2+xy\)
\(=\left(2x+3-2x+3\right)^2+xy\)
\(=xy+36=2\cdot\left(-1\right)+36=36-2=34\)
Bài 2:
a: \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng)
b: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+2ab+b^2+2ac+c^2+2bc+a^2+ab+ac+a^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
