n chẵn => n = 2k (k \(\in\)N)

n³ + 6n² + 8n = (2k)³ + 6.(2k)² + 8.(2k) = 8k³ + 24.k² + 16k = 8k. (k² + 3k + 2) = 8k.(k² + 2k + k + 2)

= 8k. [k(k +2) + (k+2)] = 8k.(k+1).(k+2)

Nhận xét: k; k+1; k+ 2 là 3 số tự nhiên liên tiếp nên tích của chúng chia hết cho 6

=>  8k.(k+1).(k+2) chia hết cho 8.6 = 48

=> n³ + 6n² + 8n chia hết cho 48

ko bk lam

4x⁴-32x²+1

=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

=2x².(2x²+6x+1)-6x.(2x²+6x+1)+(2x²+6x+1)

=(2x²+6x+1)(2x²- 6x+1)

= 4x⁴  - 4x² + 1 - 28x² = [(2x²)² - 2.2x² .1 + 1² ] - 28x² = (2x² - 1)² - (\(\sqrt{28}\).x)²

= (2x² - 1 - \(\sqrt{28}\)x) .(2x²  -1 + \(\sqrt{28}\)x)  = (2x² - 2\(\sqrt{7}\)x - 1). (2x² + 2\(\sqrt{7}\)x -1) 

(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)

= x² + 8x + 16 - 6 = (x² +2.x.4 + 4²) -  (\(\sqrt{6}\))² = (x+4)²  -  (\(\sqrt{6}\))²  = (x+4 - \(\sqrt{6}\)). (x+4+ \(\sqrt{6}\))

khó quá       

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

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)

bn Đinh Tuấn Việt có nick hoc 24 h à

(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)

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