Cho $x, y, z$ dương thoả mãn $(x^2-xy+y^2)=xy.(x+y)$
Tìm min, max của $A;$        $A=(\frac{1}{x^3}+\frac{1}{y^3}  )$
bt này đề kt lp tớ cx có –  jessica2308.snsd 07-03-16 11:17 AM
Ta có: $x^2-xy+y^2=xy(x+y)\Leftrightarrow \frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}=\frac{1}{x}+\frac{1}{y}$.
Đặt $a=\frac{1}{x},b=\frac{1}{y},\,a,b>0$.
Khi đó giả thiết trở thành: $a^2-ab+b^2=a+b$.
Ta có: $a+b=a^2-ab+b^2\ge\frac{(a+b)^2}{2}-\frac{(a+b)^2}{4}= \frac{(a+b)^2}{4} $
Suy ra: $0<a+b\le 4$ .
Ta có: $A=a^3+b^3=(a+b)^2\le 16$
Vậy Max$A=16 \Leftrightarrow a=b=2\Leftrightarrow x=y=\frac{1}{2}$.
Min$A$ không tồn tại (Inf$A=0$ ).
đáp án rất chi tiết vote cho bạn ! hyhy –  nguyentienthanhqn 20-09-12 11:44 PM
cám ơn bạn nhiều nhé –  nguyentienthanhqn 19-09-12 11:02 PM
Ta có: a+b=a2ab+b2(a+b)22(a+b)24= (a+b)24
Suy ra: 0<a+b4 .
Ta có: A=a3+b3=(a+b)216
Vậy MaxA=16 a=b=2x=y=12.
MinA không tồn tại (InfA=0 ).

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