phan tich đa thức thành nhân tử: (x+2)(x+3)(x+4)(x+5)-24
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= x2 + 8x + 16 - 6 = (x2 +2.x.4 + 42) - (\(\sqrt{6}\))2 = (x+4)2 - (\(\sqrt{6}\))2 = (x+4 - \(\sqrt{6}\)). (x+4+ \(\sqrt{6}\))
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vậy 2x3-x2+5x+3 = 0 <=> (2x+1).(x2 - x+ 3) = 0 <=> 2x+1 = 0 hoặc x2 - x + 3 = 0
+) 2x+1 = 0 <=> x = -1/2
+) x2 - x+ 3 = 0 Vô nghiệm vì x2 - x+ 3 = x2 - 2.x. \(\frac{1}{2}\)+ \(\frac{1}{4}\) + \(\frac{11}{4}\) = (x - \(\frac{1}{2}\)) 2 + \(\frac{11}{4}\)> 0 với mọi x
Vậy phương trình có nghiệm x = -1/2
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Ta có a+b+c+d=0
=> a+b=-(c+d)
=> (a+b)^3=-(c+d)^3
=> a^3+b^3+3ab(a+b) = -c^3-d^3-3cd(c+d)
=> a^3+b^3+c^3+d^3 = -3ab(a+b)-3cd(c+d)
=> a^3+b^3+c^3+d^3 = 3ab(c+d)-3cd(c+d) (vì a+b = - (c+d))
=> a^3 +b^3+c^3+d^3 = 3(c+d)(ab-cd) (đpcm)
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(a+b+c)3=(a+b)3+3(a+b)2.c+3.(a+b).c2+c3
=a3+3a2b+3ab2+b3+3(a+b)2.c+3.(a+b).c2+c3
=a3+b3+c3+[3a2b+3ab2+3(a+b)2.c+3.(a+b).c2]
=a3+b3+c3+[3ab(a+b)+3(a+b)2c+3(a+b)c2]
=a3+b3+c3+3(a+b)[ab+(a+b)c+c2]
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[a.(b+c)+c.(b+c)]
=a3+b3+c3+3(a+b)(b+c)(a+c)
=> ĐPCM
(a+b+c)3 = (a + b)3 + c3 + 3(a+b)c.(a+b+c)
= a3 + b3 + 3ab.(a+b) + c3 + 3(a+b)c(a+b+c) = a3 + b3 + c3 + 3(a+b). (ab + ac + bc + c2 )
= a3 + b3 + c3 + 3.(a+b). [a(b+c) + c.(b+c)] = a3 + b3 + c3 + 3(a+b).(a+c).(b+c)
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x=14
=>15=x+1
16=x+2
29=x+15
13=x-1
=>A=x5-15x4+16x3-29x2+13x
=x5-(x+1)x4+(x+2)x3-(x+15)x2+(x-1)x
=x5-x5-x4+x4+2x3-x3-15x2+x2-x
=x3-14x2-x
=x3-x.x2-x
=x3-x3-x
=-x
=-14
Vậy A=-14
x=14
=>15=x+1
16=x+2
29=x+15
13=x-1
=>A=x5-15x4+16x3-29x2+13x
=x5-(x+1)x4+(x+2)x3-(x+15)x2+(x-1)x
=x5-x5-x4+x4+2x3-x3-15x2+x2-x
=x3-14x2-x
=x3-x.x2-x
=x3-x3-x
=-x
=-14
Vậy A=-14
(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10 ta được:
t.(t+2)-24
=t2+2t-24
=t2-4t+6t-24
=t.(t-4)+6.(t-4)
=(t-4)(t+6)
thay t= x2+7x+10 ta được:
(x2+7x+6)(x2+7x+16)
=(x2+x+6x+6)(x2+7x+16)
=[x.(x+1)+6.(x+1)](x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
Vậy (x+2)(x+3)(x+4)(x+5)-24=(x+1)(x+6)(x2+7x+16)
= [(x+2).(x+5)]. [(x+3).(x+4)] - 24 = (x2 + 5x + 2x+ 10). (x2 + 4x+3x+12) - 24
= (x2 + 7x + 10).(x2 + 7x + 12) - 24
= (x2 + 7x + 10). [(x2 + 7x + 10)+ 2] - 24 = (x2 + 7x + 10)2 + 2. (x2 + 7x + 10) - 24
= (x2 + 7x + 10)2 + 6 (x2 + 7x + 10) - 4(x2 + 7x + 10) - 24
= [ (x2 + 7x + 10)2 + 6 (x2 + 7x + 10)] - [4(x2 + 7x + 10) + 24]
= (x2 + 7x + 10) . [(x2 + 7x + 10) + 6] - 4. [(x2 + 7x + 10) + 6]
= (x2 + 7x + 10 - 4). [(x2 + 7x + 10) + 6] = (x2 + 7x + 6). (x2 + 7x + 16)
= (x2 + x+ 6x + 6). (x2 + 7x + 16) = [x(x+1) + 6.(x+1)]. (x2 + 7x + 16) = (x+6).(x+1).(x2 + 7x + 16)