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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) \(ax+by+bx+ay=\left(ax+bx\right)+\left(by+ay\right)\)
\(=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)
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) \(ax+by+bx+ay=ax+bx+by+ay=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)
ĐẶT NHÂN TỬ CHUNG NHA!
1) ab - ac + ad = a( b- c +d )
2) ax - bx - cx + dx = x( a-b-c+d)
3) a.( b + c ) - d . ( b + c )= (b+c)(a-d)
4) ac - ad + bc - bd = a( c-d) + b( c-d) = (a+b)(c-d)
5) ax + by + bx + ay= a( x+y) + b( x+y) = (a+b)(x+y)
1, ab + ac = a(b+c)
2, ab - ac + ad = a(b-c+d)
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, ax+by+bx+ay=ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)
a(b+c)
a(b-c+d)
x(a-b-c+d)
(b+c)(a-d)
a(c-d)+c(c-d)=(c-d)(a+c)
a(x+y)+b(x+y)=(x+y)(a+b)
T
1) ab + ac = a(b + c)
2) ab - ac + ad = a(b - c + d)
3) ax - bx - cx + dx = x(a - b - c + d)
4) a(b + c) - d(b + c) = (b + c)(a - d)
5) ac - ad + bc - bd = a(c - d) + b(c - d) = (c - d)(a + b)
6) ax + by + bx + ay = (ax + ay) + (bx + by) = a(x + y) + b(x + y) = (x + y)(a + b)
ab + ac = a ( b + c )
ab - ac + ad = a ( b - c + d )
ax - bx - cx + dx = x ( a - b - c + d )
1/ab+ac=a(b+c)
2/ab-ac+ad=a(b-c)+ad=a(b-c+d)
3/ax-bx-cx+dx=x(a-b-c)+xd=x(a-b-c+d)
4/a(b+c)-d(b+c)=(ab+ac)-(bd+cd)=b(a+d)-c(a+d)=a+d(b+c)
5/ac-ad+bc-bd=a(c-d)+b(c-d)=c-d(a+b)
6/ax+by+bx+ay=a(x+y)+b(x+y)=x+y(a+b)
1?
a(b+c)
a(b-c+d)
x(a-b-c+d)
(b+c)(a-d)
(c-d)(a+b)
(x+-y)(a+b)
Câu c với f là mình sửa lại nhé, chắc gõ nhầm :v
Bài 1:
a, ab + ac
= a(b + c)
b, ab - ac + a
= a(b - c + 1)
c, ax - b - (x + ax)
= ax - b - x - ax
= -b - x
= -1(b + x)
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)
f, ax + by + bx + ay
= a(x + y) + b(x + y)
= (x + y)(a + b)
Bài 2:
a, n + 7 \(⋮\) n + 2 (n \(\ne\) -2)
n + 2 + 5 \(⋮\) n + 2
Mà n + 2 \(⋮\) n + 2
\(\Rightarrow\) 5 \(⋮\) n + 2
\(\Rightarrow\) n + 2 \(\in\) Ư(5) = {-1; 1; -5; 5}
Xét các TH:
n + 2 = -1 \(\Rightarrow\) n = -3 (TM)
n + 2 = 1 \(\Rightarrow\) n = -1 (TM)
n + 2 = -5 \(\Rightarrow\) n = -7 (TM)
n + 2 = 5 \(\Rightarrow\) n = 3 (TM)
Vậy n \(\in\) {-3; -1; -7; 3}
b, 9 - n \(⋮\) n - 3 (n \(\ne\) 3)
6 - (n - 3) \(⋮\) n - 3
Mà n - 3 \(⋮\) n - 3
\(\Rightarrow\) 6 \(⋮\) n - 3
\(\Rightarrow\) n - 3 \(\in\) Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}
Xét các TH:
n - 3 = 1 \(\Rightarrow\) n = 4 (TM)
n - 3 = -1 \(\Rightarrow\) n = 2 (TM)
n - 3 = 2 \(\Rightarrow\) n = 5 (TM)
n - 3 = -2 \(\Rightarrow\) n = 1 (TM)
n - 3 = 3 \(\Rightarrow\) n = 6 (TM)
n - 3 = -3 \(\Rightarrow\) n = 0 (TM)
n - 3 = 6 \(\Rightarrow\) n = 9 (TM)
n - 3 = -6 \(\Rightarrow\) n = -3 (TM)
Vậy n \(\in\) {4; 2; 5; 1; 6; 0; 9; -3}
c, 2n + 7 \(⋮\) n + 1 (n \(\ne\) -1)
2n + 2 + 5 \(⋮\) n + 1
2(n + 1) + 5 \(⋮\) n + 1
Ta có: n + 1 \(⋮\) n + 1 nên 2(n + 1) \(⋮\) n + 1
\(\Rightarrow\) 5 \(⋮\) n + 1
\(\Rightarrow\) n + 1 \(\in\) Ư(5) = {1; -1; 5; -5}
Xét 4 TH:
n + 1 = -1 \(\Rightarrow\) n = -2 (TM)
n + 1 = 1 \(\Rightarrow\) n = 0 (TM)
n + 1 = -5 \(\Rightarrow\) n = -6 (TM)
n + 1 = 5 \(\Rightarrow\) n = 4 (TM)
Vậy n \(\in\) {-2; 0; -6; 4}
d, 3n + 7 \(⋮\) 2n + 1 (n \(\ne\) \(\frac{-1}{2}\))
Vì 3n + 7 \(⋮\) 2n + 1 nên 2(3n + 7) \(⋮\) 2n + 1 hay 6n + 14 \(⋮\) 2n + 1
Ta có: 6n + 14 \(⋮\) 2n + 1
6n + 3 + 11 \(⋮\) 2n + 1
3(2n + 1) + 11 \(⋮\) 2n + 1
Ta có 2n + 1 \(⋮\) 2n + 1 nên 3(2n + 1) \(⋮\) 2n + 1
\(\Rightarrow\) 11 \(⋮\) 2n + 1
\(\Rightarrow\) 2n + 1 \(\in\) Ư(11) = {1; -1; 11; -11}
Xét 4 TH:
2n + 1 = 1 \(\Rightarrow\) 2n = 0 \(\Rightarrow\) n = 0 (TM)
2n + 1 = -1 \(\Rightarrow\) 2n = -2 \(\Rightarrow\) n = -1 (TM)
2n + 1 = 11 \(\Rightarrow\) 2n = 10 \(\Rightarrow\) n = 5 (TM)
2n + 1 = -11 \(\Rightarrow\) 2n = -12 \(\Rightarrow\) n = -6 (TM)
Vậy n \(\in\) {0; -1; 5; -6}
Chúc bn học tốt (Dài quá, làm mãi mới hết :v)