b.
Bunhiacopxki:
\(S\le\sqrt{3\left(5a+4+5b+4+5c+4\right)}=\sqrt{51}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Do \(\left\{{}\begin{matrix}a+b+c=1\\0\le a;b;c\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5a+4}=x\\\sqrt{5b+4}=y\\\sqrt{5c+4}=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2\le x;y;z\le3\\x^2+y^2+z^2=5\left(a+b+c\right)+12=17\end{matrix}\right.\)
Do \(2\le x;y;z\le3\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)+\left(y-2\right)\left(y-3\right)+\left(z-2\right)\left(z-3\right)\le0\)
\(\Leftrightarrow5x+5y+5z\ge x^2+y^2+z^2+18\)
\(\Leftrightarrow x+y+z\ge\dfrac{x^2+y^2+z^2+18}{5}=7\)
\(S_{min}=7\) khi \(\left(x;y;z\right)=\left(2;2;3\right)\) và hoán vị hay \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị
a.
\(\left(x+y\right)\left(x+2y\right)=x+5\)
\(\Leftrightarrow x^2+3xy+2y^2=x+5\)
\(\Leftrightarrow x^2+\left(3y-1\right)x+2y^2-5=0\) (1)
Do x;y nguyên \(\Rightarrow\Delta=\left(3y-1\right)^2-4\left(2y^2-5\right)=k^2\)
\(\Leftrightarrow y^2-6y+21=k^2\)
\(\Leftrightarrow k^2-\left(y-3\right)^2=12\)
\(\Leftrightarrow\left(k-y+3\right)\left(k+y-3\right)=12\)
\(\Rightarrow y=\left\{1;5\right\}\)
Lần lượt thế vào (1) để tìm x nguyên tương ứng
