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Bài 17 :
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 = a( x+ y ) + b ( x + y ) = ( x + y ) (a +b )
Bài 18:
1/ (a – b + c) – (a + c) = a - b + c - a - c = -b
2/ (a + b) – (b – a) + c = a + b - b + a + c = 2a + 2
3/ - (a + b – c) + (a – b – c) = -a -b + c + a - b - c = -2b
4/ a(b + c) – a(b + d) = a ( b + c - b - d ) = a( c - d )
5/ a(b – c) + a(d + c) = a ( b - c + d + c ) = a ( b+ d )
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)
...............
1 a(b+c)
2 a(b-c+d)
3 x(a-b-c+d)
4 (b+c)(a-d)
5 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)
làm được nhiu ây thui, mí bài kia tự làm nhak
hihhhi
bài 2 \
1 (a-b+c)-(a+c)=-b
phá ngoặc
=a-b+c-a-c
=-b
2 làm giống bài 1 í. phá ngoặc hớt, mí bài còn lại cũng lm tương tự
phá ngoặc là được thui :)))))
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) \(ax+by+bx+ay=ax+bx+by+ay=x\left(a+b\right)+y\left(a+b\right)=\left(a+b\right)\left(x+y\right)\)
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) ax(b+c)
b) ax(b-c+d)
c) X x(a-b-c+d)
e, f tương tự
k mình nha bạn
1.
1) a ( b + c )
2) a ( b-c+d)
3) x ( a-b-c+d)
4) ( b+c ) (a - d )
5) a (c-d) + b (c-d) =(c-d) (a + b )
6) a ( x+y) + b ( y+x) = (x+y) ( a+b)
2.
1) a - b + c - a - c = -b
2) a + b - b + a + c = 2a + c
3) - a - b + c + a - b - c = -2b
4) ab + ac - ab - ad = ac-ad = a (c-d)
5) ab - ac + ad + ac = ab + ad = a (b+d)
ab + ac = a(b + c)
ab - ac + ad = a(b - c + d)
ax - bx - cx + dx
=x(a - b - c + d)
\(a,\left(a-b+c\right)-\left(a+c\right)=-b\)
Ta có: \(VT=\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b=VP\left(đpcm\right)\)
\(b,\left(a+b\right)-\left(b-a\right)+c=2a+c\)
Ta có: \(VT=\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c=VP\left(đpcm\right)\)
\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-2b\)
Ta có: \(VT=-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b=VP\left(đpcm\right)\)
\(d,a\left(b+c\right)-a\left(b+d\right)=a\left(c-d\right)\)
Ta có: \(VT=a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)=VP\left(đpcm\right)\)
\(e,a\left(b-c\right)+a\left(d+c\right)=a\left(b+d\right)\)
Ta có: \(VT=a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)=VP\left(đpcm\right)\)
Bài 1:
a, (a - b + c) - (a + c)
= a - b + c - a - c
= ( a - a ) + ( c- c ) - b
= -b
b, ( a + b ) - (b - a ) + c
= a + b - b + a + c
= ( a + a ) + ( b- b ) + c
= 2a + c
c, - ( a + b - c) + (a - b -c)
= -a -b + c + a - b - c
= ( -a + a ) - ( b + b ) + ( c - c)
= - 2b
d, a(b + c) - a ( b + d )
= ab + ac - ab - ad
= ac - ad
= a( c-d )
e, a( b - c ) + a( d + c )
= ab - ac + ad + ac
= ab + ad
= a( b+d )
Bài 2:
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) = (a+b).( c - d )
g, ax + by + bx + ay = a( x + y) + b(y + x) = ( a+b).( x+y )