Question

Cho \(A=a\left(b-c\right)-b\left(a-c\right)\) và \(B=b\left(a-c\right)-a\left(b+c\right)\)

Tính tổng A + B và hiệu A - B

Answer 1

Ta có :

A = a.(b - c) - b.(a - c)

A = a.b - a.c - b.a + b.c

A = (a.b - b.a) + [-(a.c) + b.c]

A = 0 + [c.(-a + b)]

A = c.(-a + b)

B = b.(a - c) - a.(b + c)

B = b.a - b.c - a.b + a.c

B = (b.a - a.b) + [-(b.c) + a.c]

B = 0 + [c.(-b + a)]

B = c.(-b + a)

\(\Rightarrow\) A + B = c(-a + b) + c(-b +a)

= c(-a + b - b + a)

= c . 0

= 0

\(\Rightarrow\) A - B = c.(-a + b) - c(b + a)

= c(-a + b - b - a)

= c . (-2a)

Answer 2

\(A+B=a\left(b-c\right)-b\left(a-c\right)+b\left(a-c\right)-a\left(b+c\right)\\ =\left(ab-ac\right)-\left(ab+bc\right)+\left(ab-bc\right)-\left(ab+ac\right)\\ =ab-ac-ab-bc+ab-bc-ab-ac\\ =0\)

\(A-B=a\left(b-c\right)-b\left(a-c\right)-\left(b\left(a-c\right)-a\left(b+c\right)\right)\\ =ab-ac-ab+bc-\left(ab-bc-ab-ac\right)\\ =ab-ac-ab+bc-ab+bc+ab+ac\\ =2.bc\)