Question

Tính:

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

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

Answer 1

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

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

Answer 2

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Answer 3

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

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

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

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

Answer 4

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

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

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

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

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

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

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

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

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

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

Answer 5

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

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

Answer 6

\(\text{a)}\) \(\left(a+b+c\right)^2\)

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

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

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

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

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

\(\text{b)}\) \(\left(a-b+c\right)^2\)

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

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

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

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

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