\(\left(x^2+9\right)\left(x+3\right)\left(x-3\right)-\left(x^2-3\right)^2=\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-9\right)\)

\(=x^4-81-x^4+9=-72\)

a) Co: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) ( vi a+b = - (c+d))
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd)                             (dpcm)

b) Co: a+b+c=9

=> (a+b+c)^2 = 49

=> a^2 + b^2 +c^2 + 2(ab + bc + ca)  = 49

=> 2(ab+bc+ca) = -4

=> ab+bc+ca= -2

2) \(8x^3-12x^2+6x-1=0\leftrightarrow\left(2x-1\right)^3=0\leftrightarrow2x-1=0\leftrightarrow x=\frac{1}{2}\)
 

= \(2x^3\)+\(2\)

ai giup mk vs huhu

ai giup mk cau c) mk kich cho

a) A=-(x²-4x-7)=\(-\left[\left(x-2\right)^2-11\right]=-\left(x-2\right)^2+11\)

ta có -(x-2)² \(\le\)0

-> -(x-2)²+11 \(\le\)11

--> A​\(\le\)​11

vậy GTLN của A là 11

b) B= - (x²-5x+127)= - (x-5/2)²-483/4

c) C= - (x²+4x) = - (x+2)²+4

Ta có M=N=P=Q=1:2:3:4 => \(\frac{M}{1}\)= \(\frac{N}{2}\)=\(\frac{P}{3}\)=\(\frac{Q}{4}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :  \(\frac{M}{1}\)= \(\frac{N}{2}\)=\(\frac{P}{3}\)=\(\frac{Q}{4}\)= \(\frac{M+N+P+Q}{1+2+3+4}\)=\(\frac{360}{10}\)= 36

Suy ra  \(\frac{M}{1}\)= 36 => M=36 ; N/2 = 36 => N=36.2=72; P/3=36=>P=36.3=108; Q/4=36=>Q=36.4=144

A= x(x+5)+3(x+5)+4 =x²+5x+3x+15+4 =x²+8x+19 =x²+2.4.x+16+3=(x+4)²+3

ta thay : (x+4)²>hoac = 0 suy ra Amin khi va chi khi x+4=0 suy ra x=-4 

Vay Amin = 3 khi x=-4

 B=x²-4x+4+y²-8y+16-14 =(x-2)²+(y-4)²-14

vi (x-2)²  va (y-4)² lon hon hoac bang 0 suy ra Bmin khi va chi khi (x-2)²=0 va (y-4)²=0

tinh ra nhu cau a (ban tu lam nhe)

vay Bmin=-14 va x=2 va y=4

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