Tìm số nguyên n:

a) n² + n + 17 ⋮ n + 1

b) n² + 25 ⋮ n + 2

c) 3n² + 5 ⋮ n - 1

d) 2n² + 11 ⋮ 3n + 1

là Lđó

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

Câu 1: So sánh P và Q

P = a + {(a – 3) – [( a + 3) – (- a – 2)]}.

Q = [ a + (a + 3)] – [( a + 2) – (a – 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/ 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)