1) ta có: A= x^3 -8y^3=> A=(x-2y)(x^2 +2xy+4y^2)=>A=5.(29+2xy)   (vì x-2y=5 và x^2+4y^2=29)     (1)

Mặt khác : x-2y=5(gt)=> (x-2y)^2=25=> x^2-4xy+4y^2=25=>29-4xy=25(vì x^2+4y^2=29)

                                                                                          => xy=1    (2)

Thay (2) vào (1) ta đc: A= 5.(29+2.1)=155

Vậy gt của bt A là 155

2) theo bài ra ta có: a+b+c=0 => a+b=-c=>(a+b)^2=c^2=> a^2 +b^2+2ab=c^2=>c^2-a^2-b^2=2ab

=> \(\left(c^2-a^2-b^2\right)^2=4a^2b^2\)

=>\(c^4+a^4+b^4-2c^2a^2+2a^2b^2-2b^2c^2=4a^2b^2\)

=>\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

=> \(a^4+b^4+c^4=\frac{1}{2}\left(a^2+b^2+c^2\right)^2\) (đpcm)

vì a,b,c là 3 số thực khác nhau và khác 0 nên a-b, b-c, a-c khác 0. Do đó:

a²- b= b²- c <=> a² -b² =b -c <=>(a-b)(a+b)=b-c => a+b =(b-c)/(a-b)

cmtt ta có b+c=(c-a)/(b-c) ; c+a = (a-b)/(c-a). Như vậy ta tính được P=1

a+b=1

=>b=1-a; a=1-b

\(b^3-1=\left(b-1\right)\left(b^2+b+1\right)\)

=-a(b^2+b+1)

\(a^3-1\) =(a-1)\(\left(a^2+a+1\right)\)

=-b(a^2+a+1)

\(\frac{a}{b^3-1}-\frac{b}{a^3-1}\)

\(=\frac{a}{-a\left(b^2+b+1\right)}-\frac{b}{-b\left(a^2+a+1\right)}\)

\(=\frac{-1}{b^2+b+1}+\frac{1}{a^2+a+1}\)

\(=\frac{(b^2 + b + 1) - (a^2 + a + 1)}{(a^2 + a + 1)(b^2 + b + 1)}\)

\(=\frac{\left(b^2-a^2\right)+\left(b-a\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}=\frac{\left(b-a\right)\left(b+a+1\right)}{\left(a^2+a+1\right)\left(b^2+b+1\right)}\)

\(=\frac{2\left(b-a\right)}{a^2b^2+a^2b+a^2+ab^2+ab+a+b^2+b+1}\)

\(=\frac{-2\left(a-b\right)}{a^2b^2+ab(a+b)+ab+(a+b)^2-2ab+(a+b)+1}\)

\(=\frac{-2\left(a-b\right)}{=a^2b^2+ab\cdot1+ab+1^2-2ab+1+1}=\frac{-2\left(a-b\right)}{a^2b^2+3}\)

=>\(P=\frac{a}{b^3-1}-\frac{b}{a^3-1}+\frac{2\left(a-b\right)}{a^2b^2+3}=\frac{-2\left(a-b\right)+2\left(a-b\right)}{a^2b^2+3}\) =0

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)

Lời giải:

\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)

\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)

\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)

\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)