Câu 12: Chứng minh đẳng thức
1/ (a – b + c) – (a + c) = -b
2/ (a + b) – (b – a) + c = 2a + c
3/ – (a + b – c) + (a – b – c) = -2b
4/ a(b + c) – a(b + d) = a(c – d)
5/ a(b – c) + a(d + c) = a(b + d)
6/ a.(b – c) – a.(b + d) = -a.( c + d)
7/ (a + b).( c + d) – (a + d).( b + c) = (a – c). (d – b)
1/ (a – b + c) – (a + c) = -b
VT = a – b + c – a – c = -b
⇒ đpcm
2/ (a + b) – (b – a) + c = 2a + c
VT = a + b – b + a + c = 2a + c
⇒ đpcm
3/ – (a + b – c) + (a – b – c) = -2b
VT = -a – b + c + a -b – c = = -2b
⇒ đpcm
4/ a(b + c) – a(b + d) = a(c – d)
VT = ab + ac – ab – ad = a( c-d)
⇒ đpcm
5/ a(b – c) + a(d + c) = a(b + d)
VT = ab – ac + ad + ac = a(b+d)
⇒ đpcm
6/ a.(b – c) – a.(b + d) = -a.( c + d)
VT = ab – ac -ab – ad = -a(c+d)
⇒ đpcm
7/ (a + b).( c + d) – (a + d).( b + c) = (a – c). (d – b)
VT = ac + ad + bc + bd – ( ab + ac + bd + cd )
= ac + ad + bc + cd – ab – ac – bd – cd
= ( ac – ac ) + ( bd – bd ) + ad + bc – ab -cd
= a(d-b) – c(d-b)
= ( a- c)(d-b)
⇒ Đpcm
1) (a – b + c) – (a + c) = -b
Xét VT: (a – b + c) – (a + c) = a -b +c -a -c
= (a -a) + (c-c) -b
= -b = VP
⇒ ĐPCM
2) (a + b) – (b – a) + c = 2a + c
Xét VT: (a + b) – (b – a) + c = a +b -b +a +c
= (a +a) + (b-b) +c
= 2a +c = VP
⇒ ĐPCM
3) – (a + b – c) + (a – b – c) = -2b
Xét VT: – (a + b – c) + (a – b – c) = -a -b +c +a -b -c
= ( -a+a) – (b+b) + (c-c)
= -2b = VP
⇒ ĐPCM
4) a(b + c) – a(b + d) = a(c – d)
Xét VT: a(b + c) – a(b + d) = ab +ac -ab -ad
= (ab -ab) + a(c -d)
= a.(c-d) = VP
⇒ ĐPCM
5) a(b – c) + a(d + c) = a(b + d)
Xét VT: a(b – c) + a(d + c) = ab -ac +ad +ac
= ( -ac +ac) + a(b+d)
= a( b+d) = VP
⇒ ĐPCM
6) a.(b – c) – a.(b + d) = -a.( c + d)
Xét VT: a.(b – c) – a.(b + d) = ab – ac -ab -ad
= (ab -ab) – a(c +d)
= -a.(c+d) = VP
⇒ ĐPCM
7) (a + b).( c + d) – (a + d).( b + c) = (a – c). (d – b)
Xét VT: (a + b).( c + d) – (a + d).( b + c) = ac +ad +bc +bd -ab -ac -bd -cd
= (ac -ac) + (bd-bd) +ad -ab -cd +bc
= a(d-b) – c(d-b)
= (d-b).(a-c) = VP
⇒ ĐPCM