bai1:viet duoi dang tich cac tong sau
a)ab+ac
b)ab-ac+ad
c)ax-bx-d(b+c)
d)ac-ad+bc-bd
f)a(b+c)-d(b+c)
e)ax+by+bx+ay
BAI 1 viet duoi dang tich cua cac tong sau
1,ab+ac
2, ab- ac +a
3, ax - bx -cx +dx
4,a nhan (b+c ) -d (b +c)
5, ac _ ad + bc - bd
6, ax +by +bx+ay
\(ab+ac=a\left(b+c\right)\)
\(ab-ac+a=a\left(b-c+1\right)\)
\(ax-bx-cx+dx=x\left(a-b-c-d\right)\)
\(a\left(b+c\right)-d\left(b+c\right)=0\Leftrightarrow\left(b+c\right)\left(a-d\right)\)
\(ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(a+b\right)\left(c-d\right)\)
\(ax+by+bx+ay=\left(ax+ay\right)+\left(bx+by\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
ab + ac = a( b + c )
ab - ac + a = a( b - c +1 )
ax - bx - cx + dx = x( a - b - c - d )
a( b + c ) - d( b + c ) = 0 <=> ( b + c ) ( a - d )
ac - ad + bc - bd = a( c - d ) + b( c - d ) = ( a + b ) ( c - d )
ax + by + bx + ay = ( ax + ay ) + ( bx + by ) = a( x + y ) + b( x + y ) = ( a + b ) ( x + y )
1, ab + ac = a . ( b + c )
2, ab - ac + a = a . ( b - c + 1)
3, ax - bx - cx + dx = x . ( a - b - c + d )
4, a . ( b + c ) - d . ( b + c) = ( a - d ) . ( b + c )
5, ac - ad + bc - bd = a . ( c - d ) + b . ( c - d ) = ( a + b ) . ( c - d )
6, C1: ax + by + bx + ay = ( ax + ay ) + ( bx + by ) = a . ( x + y ) + b . ( x + y ) = ( a + b ) . ( x + y)
C2: ax + by + bx + ay = ( ax + bx ) + ( ay + by ) = x . ( a + b ) + y . ( a + b ) = ( x + y ) . (a + b )
Viết ưới dạng tích các tổng sau:
a, ab + ac
b, ab - ac + ad
c, ax - bx - cx + dx
d, a( b + c ) - d( b + c )
e, ac - ad + bc - bd
f, ax + by + bx + ay
a) ax(b+c)
b) ax(b-c+d)
c) X x(a-b-c+d)
e, f tương tự
k mình nha bạn
Viết dưới dạng tích các tổng sau:
a) ab+ac
b) ab-ac+ad
c) ax-bx-cx+dx
d) a(b+c)-d(b+c)
e) ac-ad+bc-bd
g) ax+by+bx+ay
a) ab+ac=a(b+c)
b) ab-ac+ad=a(b-c+d)
c) ax-bx-cx+dx = x(a-b-c+d)
d) a(b+c)-d(b+c)= (b+c)(a-d)
e) ac-ad+bc-bd = a(c-d)+b(c-d)= (c-d)(a+b)
g) ax+by+bx+ay= x(a+b)+y(a+b)=(x+y)(a+b)
a,a.(b+c)
b,a.(b-c+d)
c,x.(a-b-c+d)
d,(a-d).(b+c)
...............
nhớ cho mk nha, mk làm đây, đừng cho ai khác nha, mk sẽ làm ngay bây giờ nhưng khi mk làm xong thì cho mk là đc
Viết dưới dạng tích
a) ab + ac
b) ab - ac + ad
c) ax - bx - cx + dx
d) a(b + c) - d(b + c)
e) ac - ad + bc - bd
g) ax + by + bx +ay
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, \(a\left(b+c\right)-d\left(b+c\right)\)
\(=\left(b+c\right)\left(a-d\right)\)
\(a)ab+ac=a\left(b+c\right)\)
\(b)ab-ac+ad=a\left(b-c+d\right)\)
\(c)ax-bx-cx+dx=x\left(a-b-c+d\right)\)
\(d)a\left(b+c\right)-d\left(b+c\right)=\left(b+c\right)\left(a-d\right)\)
\(e)ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(a+b\right)\left(c-d\right)\)
\(g)ax+by+bx+ay=\left(ax+bx\right)+\left(ay+by\right)=x\left(a+b\right)+y\left(a+b\right)=\left(x+y\right)\left(a+b\right)\)
Viết dưới dạng tích các tổng sau : ab+ac ; ab -ac+ad; ax -bx-cx+dx; a(b+c) - d (b+c); ac-ad+bc-bd; ax+by+bx+ay
ab + ac = a(b + c)
ab - ac + ad = a(b - c + d)
ax - bx - cx + dx
=x(a - b - c + d)
ab+ac=a(b+c)
ab-ac+ad=a(b-c+d)
ax-bx-cx+dx=x(a-b-c+d)
Viết dưới dạng các tích của tổng
a) ab + ac
b) ab - ac + ad
c) ax - bx - cx + dx
d) a(b+c) - d(b+c)
e) ac - ad + bc - bd
f) ax + by + bx + ay
a,ab+ac=a(b+c)
b,ab-ac+ad=a(b-c+d)
c,ax-bx-cx+dx=(a-b-c+d)x
d,a(b+c)-d(b+c)
=ab+ac-bd+cd
=b(a-d)+(a+d)c
e,ac-ad+bc-bd
=c(a+b)-(a-b)d
f,ax+by+bx+ay
=a(x+y)+b(y+x)
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) \(ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(c-d\right)\left(a+b\right)\)
f) \(ax+by+bx+ay=ax+bx+by+ay=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)
ab+ac
ab-ac+ad
ax-bx-cx-dx
a(b+c)-d(b+c)
ac-ad+bc-bd
ax+by+bx+ay
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!!!!!
Bài 1: Viết dưới dạng tích các tổng sau:
a/ ab + acbx
b/ ab - ac + ad
c/ ax - bx - cx + dx
d/ a(b + c) - d(b + c)
e/ ac - ad + bc - bd
f/ ax + by + bx + ay
a) \(ab+acbx=ab\left(1+cx\right)\)
b) \(ab-ac+ad=a\left(b-c+d\right)\)
c) \(ax-bx-cx+dx=x\left(a-b-x+d\right)\)
d) \(a\left(b+c\right)-d\left(b+c\right)=\left(b+c\right)\left(a-d\right)\)
e) \(ac-ad+bc-bd=a\left(c-d\right)+b\left(c-d\right)=\left(a+b\right)\left(c-d\right)\)
f) \(ax+by+bx+ay=\left(ax+bx\right)+\left(by+ay\right)\)
\(=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)
ab+ac
ab-ac+ad
ax-bx-cx +dx
a(b+c) - d(b+c)
ac-ad+bc-bd
ax+by+bx+ay
ai đúng mình tick cho
,bố thằng nào làm đc
bạn đừng vào bình luận linh tinh