a) Ta có \(ab-ac+ad=a.\left(b-c+d\right)\)
b) Ta có \(ac+ad-bc-bd=a.\left(c+d\right)-b.\left(c+d\right)\)
\(=\left(c+d\right).\left(a-b\right)\)
a) ab - ac + ad
= a.b - a.c + a.d
= a.(b-c+d)
b) ac + ad - bc - bd
= a.c + a.d - b.c - b.d
= a.(c+d) - b.(c-d)
= a.(c-d) + b.(c-d)
= (a+b) . (c-d)
a, ab - ac + ad
= a . b - a . c - a . d
= a . ( b - c - d )
b, ab + ad - bc - bd
= a . b + a . d - b . c - b .d
= a . ( b + d ) - b . ( c - d )
Đúng thì tíck mk nha !!
1)ab-ac+ad=a*(b-c+d)
2)ac+ad-bc-bd=a*(c+d)-b*(c+d)=(a-b)*(c+d)
\(a)\)
ab - ac + ad = \(a\cdot(b-c+d)\)
\(b)\)
ac + ad - bc - bd = \(a\cdot(c+d)-(bc+bd)\)
= \(a\cdot(c+d)-b\cdot(c+d)\)
\(=(c\cdot d)\cdot(a-b)\)
a ) ab - ac + ad
= a . b - a . c + a . d
= a . ( b - c + d )
b ) ac + ad - bc - bd
= a . c + a . d - b . c - b . d
= a . ( c + d ) - b . ( c - d )
= a . ( c - d ) + b . ( c - d )
= ( a + b ) . ( c - d )
a)ab-ac+ad
=a.(b-c+d)
b))ac+ad-bc-bd
=a.(c+d)-b.(c+d)
=(c+d).(a-b)
a) \(ab-ac+ad=a.b-a.c+a.d\). Đặt thừa số chung a. Ta có:
\(\Leftrightarrow a\left(b-c+d\right)\)
b) \(ac+ad-bc-bd=a.c+a.d-b.c-b.d\)
\(=a\left(c+d\right)-b\left(c-d\right)\)
\(=\left(c.d\right)+\left(a-b\right)\)
Cho sửa lại:
a) ab - ac + ad
= a.b - a.c + a.d
= a.(b-c+d)
b) ac + ad - bc - bd
= a.c + a.d - b.c - b.d
= a.(c+d) - b(c-d)
= a.(c-d) + b(c-d)
= ( a + b ) . ( c - d )