Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài

Những câu hỏi liên quan
WoflGang
Xem chi tiết
l҉o҉n҉g҉ d҉z҉
8 tháng 9 2020 lúc 16:33

A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )

= 4x( ac + bc + a + b )

= 4x[ c( a + b ) + ( a + b ) ]

= 4x( a + b )( c + 1 )

B = ax - bx + cx - 3a + 3b - 3c

= x( a - b + c ) - 3( a - b + c )

= ( a - b + c )( x - 3 )

C = 2ax - bx + 3cx - 2a + b - 3c

= x( 2a - b + 3c ) - ( 2a - b + 3c )

= ( 2a - b + 3c )( x - 1 )

D = ax - bx - 2cx - 2a + 2b + 4c

= x( a - b - 2c ) - 2( a - b - 2c )

= ( a - b - 2c )( x - 2 )

E = 3ax2 + 3bx2 + ax + bx + 5a + 5b

= 3x2( a + b ) + x( a + b ) + 5( a + b )

= ( a + b )( 3x2 + x + 5 )

F = ax2 - bx2 - 2ax + 2bx - 3a + 3b

= x2( a - b ) - 2x( a - b ) - 3( a - b )

= ( a - b )( x2 - 2x - 3 )

= ( a - b )( x2 + x - 3x - 3 )

= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]

= ( a - b )( x + 1 )( x - 3 )

Khách vãng lai đã xóa
Thanh Vu
Xem chi tiết
Truy kích
21 tháng 7 2017 lúc 21:47

bài 2:

\(A=\left(a+b+c\right)^3+\left(b+a-c\right)^3+\left(c+a-b\right)^3\)

\(=\left(c+b+a-2c\right)^3+\left(c+a+b-2b\right)^3\)

\(=\left(-2c\right)^3+\left(-2b\right)^3=-8\left(b+c\right)\)

sao nữa nhỉ :v

Shizuka
Xem chi tiết
Nguyễn Trọng Tân
Xem chi tiết
I don
19 tháng 2 2018 lúc 14:37

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!!!!!

Mael Thụy Kha
Xem chi tiết
Trần Tuyết Như
16 tháng 7 2016 lúc 14:38

a/ \(ab-2b-3a+6=\left(ab-2b\right)-\left(3a-6\right)=b\left(a-2\right)-3\left(a-2\right)=\left(a-2\right)\left(b-3\right)\)

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

c/ \(ax+by-ay-bx=\left(ax-ay\right)+\left(by-bx\right)=a\left(x-y\right)+b\left(y-x\right)=a\left(x-y\right)-b\left(x-y\right)=\left(x-y\right)\left(a-b\right)\)

d/ \(a^2-\left(b+c\right)a+bc=a^2-ab-ac+bc=\left(a^2-ac\right)+\left(ab-bc\right)=a\left(a-c\right)+b\left(a-c\right)=\left(a-c\right)\left(a+b\right)\)e/ \(\left(3a-2\right)\left(4a-3\right)-\left(2-3a\right)\left(3a+1\right)=\left(3a-2\right)\left(4a-3\right)+\left(3a-2\right)\left(3a+1\right)=\left(3a-2\right)\left(4a-3+3a+1\right)=\left(3a-2\right)\left(7a-2\right)\)

f/ \(ax+ay+az-bx-by-bz-x-y-z=\left(ax+ay+az\right)-\left(bx+by+bz\right)-\left(x+y+z\right)\)

\(=a\left(x+y+z\right)-b\left(x+y+z\right)-\left(x+y+z\right)=\left(x+y+z\right)\left(a-b-1\right)\)

Uyên Thảo
Xem chi tiết
💋Bevis💋
17 tháng 7 2019 lúc 12:18

\(a,xy+1-x-y\)

\(=\left(xy-y\right)+\left(1-x\right)\)

\(=y\left(x-1\right)- \left(x-1\right)\)

\(=\left(x-1\right)\left(y-1\right)\)

\(b,ax+ay-3x-3y\)

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

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

\(c,x^3-2x^2+2x-4\)

\(=x^2\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x^2+2\right)\left(x-2\right)\)

\(d,x^2+ab+ax+bx\)

\(=\left(x^2+ax\right)+\left(ab+bx\right)\)

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

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

\(e,16-x^2+2xy-y^2\)

\(=4^2-\left(x^2-2xy+y^2\right)\)

\(=4^2-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

💋Bevis💋
17 tháng 7 2019 lúc 12:22

\(f,ax^2+ax-bx^2-bx-a+b\)

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

\(=x^2\left(a-b\right)+x\left(a-b\right)-\left(a-b\right)\)

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

Uyên Thảo
Xem chi tiết
Cao Nguyệt Hạnh
Xem chi tiết
Nhật Hạ
6 tháng 2 2019 lúc 14:43

a, \(ab+ac\)

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

b, \(ab-ac+ad\)

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

Nhật Hạ
6 tháng 2 2019 lúc 14:44

c, \(ax-bx-cx+dx\)

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

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

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

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

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

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

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

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

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

Uyên Thảo
Xem chi tiết
Ngân Vũ Thị
17 tháng 7 2019 lúc 10:51

Hỏi đáp Toán

Trần Thanh Phương
17 tháng 7 2019 lúc 10:52

a) \(xy+1-x-y\)

\(=x\left(y-1\right)-\left(y-1\right)\)

\(=\left(y-1\right)\left(x-1\right)\)

b) \(ax+ay-3x-3y\)

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

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

c) \(x^3-2x^2+2x-4\)

\(=x^2\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+2\right)\)

d) \(x^2+ab+ax+bx\)

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

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

e) \(16-x^2+2xy-y^2\)

\(=16-\left(x^2-2xy+y^2\right)\)

\(=4^2-\left(x-y\right)^2\)

\(=\left(4-x+y\right)\left(4+x-y\right)\)

f) \(ax^2+ax-bx^2-bx-a+b\)

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

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

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

Nguyễn Thị Diễm Quỳnh
17 tháng 7 2019 lúc 10:58

a) \(xy+1-x-y=\left(xy-x\right)+\left(1-y\right)=x\left(y-1\right)+\left(1-y\right)=x\left(y-1\right)-\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)

b) \(ax+ay-3x-3y=a\left(x+y\right)-3\left(x-y\right)=\left(a-3\right)\left(x+y\right)\)

c) \(x^3-2x^2+2x-4=x^2\left(x-2\right)+2\left(x-2\right)=\left(x^2+2\right)\left(x-2\right)\)

d) \(x^2+ab+ax+bx=\left(x^2+ax\right)+\left(ab+bx\right)=x\left(x+a\right)+b\left(a+x\right)=\left(x+b\right)\left(x+a\right)\)

e) \(16-x^2+2xy-y^2=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

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