Question

Viết các biểu thức sau dưới dạng tổng của ba bình phương:

(a+b+c)^2+a^2+b^2+c^2

2(a-b)(c-b)+2(b-a)(c-a)+2(b-c)(a-c)

Answer 1

1 : \(\left(a+b+c\right)^2+a^2+b^2+c^2\)

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

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

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

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

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

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

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

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

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

Answer 2

\(\)\(1.\) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)

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

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

\(\Leftrightarrow\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)

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

\(\Leftrightarrow2ac-2ab-2bc+2b^2+2bc-2ab-2ac+2a^2+2ab-2bc-2ac+2c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\)

Tới đây dùng HĐT