Question

1) ghpt \(\left\{{}\begin{matrix}\sqrt[3]{x}+\sqrt[3]{y}=1\\\sqrt[3]{x-1}+\sqrt[3]{y+1}=1\end{matrix}\right.\)

2) cho a,b \(\ge\)0 thỏa mãn \(a^2+b^2=1\)

tìm Min lẫn Max của P= \(\sqrt{2a+1}+\sqrt{2b+1}\)

Answer 1

1) hệ <=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}\left(\sqrt[3]{x-1}+\sqrt[3]{y+1}\right)=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=1\end{matrix}\right.\)

trừ vế theo vế => \(3\sqrt[3]{xy}-3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=0\)

<=> xy=(x-1)(y-1) <=> x-y=1=> \(\left\{{}\begin{matrix}\sqrt[3]{x}+\sqrt[3]{y}=1\\x-y=1\end{matrix}\right.\)

đặt \(\sqrt[3]{x}=a;\sqrt[3]{y}=b\)

hpt <=> \(\left\{{}\begin{matrix}a+b=1\\a^3-b^3=1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}b=1-a\\2a^3-3a^2+3a-2=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}b=1-a\\\left(a-1\right)\left(2a^2-a+2\right)=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

p/s: cách làm khá dài ,có ai có cách khác thì làm luôn cho mik exp :v )

Answer 2

câu 2) xét \(p^2\) nhé ( để mai làm đã @@ )