Question

[NGUYỄN MAI HUYỀN]

Bài 1 : Tính

1) a . ( b - c ) + b . ( c - a ) + c . ( a - b )

2 ) a . ( bz - cy ) + b . ( cx - az ) + c . ( ay - bx )

Bài 2 . Chứng minh hằng đẳng thức

\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}=\dfrac{x+b}{3b-a}\)

Answer 1

[Trần Minh Hoàng]

Bài 1:

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

\(=ab-ac+bc-ba+ca-cb\)

\(=0\)

2) \(a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)\)

\(=abz-acy+bcx-baz+cay-cbx\)

\(=0\)

Answer 2

[Trần Minh Hoàng]

Bài 2:

Ta có:

\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)

\(=\dfrac{\left(x^2+bx\right)+\left(ax+ab\right)}{\left(3bx-ax\right)+\left(3ab-a^2\right)}\)

\(=\dfrac{x\left(x+b\right)+a\left(x+b\right)}{x\left(3b-a\right)+a\left(3b-a\right)}\)

\(=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}\)

\(=\dfrac{x+b}{3b-a}\)