\(A=\frac{ac+bx+ax+bc}{ay+2bx+2ax+by}=\frac{a\left(c+x\right)+b\left(c+x\right)}{a\left(y+2x\right)+b\left(y+2x\right)}=\frac{\left(c+x\right)\left(a+b\right)}{\left(y+2x\right)\left(a+b\right)}\)

Do \(a\ne-b\Rightarrow a+b\ne0\Rightarrow\)\(A=\frac{c+x}{y+2x}\), giá trị không phụ thuộc vào a; b (đpcm)

a) A = ax - ay - bx + by 

A = a(x - y) - b(x - y)

A = (a - b)(x - y)

A = 1. (-2) = -2

B = 2ax + 2ay - 2bx - 2by

B = 2a(x + y) - 2b(x + y)

B = 2(x + y)(a - b)

B = 2. (-5).(-2) = 20

a) \(ab+ac=a.\left(b+c\right)\)

b) \(ab-ac+ad=a.\left(b-c+d\right)\)

c) \(ax-bx-cx-dx=x.\left(a-b-c-d\right)\)

d) \(a.\left(b+c\right)-d.\left(b+c\right)=ab+ac-db-dc=b.\left(a-d\right)+c.\left(a-d\right)=\left(a-d\right).\left(b+c\right)\)

e) \(ac-ad+bc-bd=a.\left(c-d\right)+b.\left(c-d\right)=\left(c-d\right).\left(a+b\right)\)

f) \(ax+by+bx+ay=a.\left(x+y\right)+b.\left(y+x\right)=\left(x+y\right).\left(a+b\right)\)

CHÚC BN HỌC TỐT!!!!!

ab + ac = a(b + c)

ab - ac + ad = a(b - c + d)

ax - bx - cx + dx

=x(a - b - c + d)

chuẩn men

ab+ac=a(b+c)

ab-ac+ad=a(b-c+d)

ax-bx-cx+dx=x(a-b-c+d)

,bố thằng nào làm đc

 bạn đừng vào bình luận linh tinh

kh trả lời thì thôi

a, \(ab+ac\)

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

b, \(ab-ac+ad\)

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

c, \(ax-bx-cx+dx\)

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

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

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

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

\(b)ab-ac+ad=a\left(b-c+d\right)\)

\(c)ax-bx-cx+dx=x\left(a-b-c+d\right)\)

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

\(e)ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(a+b\right)\left(c-d\right)\)

\(g)ax+by+bx+ay=\left(ax+bx\right)+\left(ay+by\right)=x\left(a+b\right)+y\left(a+b\right)=\left(x+y\right)\left(a+b\right)\)

a) \(ay+by+ax+bx\)

\(=y\left(a+b\right)+x\left(a+b\right)\)

\(=\left(a+b\right)\left(x+y\right)\)

b) \(3xy-6y+2ax-4a\)

\(=3y\left(x-2\right)+2a\left(x-2\right)\)

\(=\left(x-2\right)\left(3y+2a\right)\)

c) sai đề ?

p/s: Trần Quỳnh Phương tên đẹp đấy nhưng lần sau nhớ viết đề cẩn thân hơn nhé!

\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)

\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)

\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)

\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)