= 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

x=-1/6            

bn viết lại đề đi 

2(a³ + b³) - 3(a² + b²) = 2(a + b)(a² - ab + b²) - 3a² - 3b²

= 2.1.(a² - ab + b²) - 3a² - 3b²

= 2a² - 2ab + 2b² - 3a² - 3b²

= - a² - b² - 2ab

= - (a + b)²

= - 1