Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left|x-6\right|+\left|-y-1\right|\ge\left|x-y-7\right|=\left|\sqrt{3}-7\right|=-\sqrt{3}+7\)
\(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=7\end{matrix}\right.\) \(\Rightarrow S=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left|2x+1\right|+\left|-2y-4\right|\ge\left|2\left(x-y\right)-3\right|=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-2\end{matrix}\right.\) \(\Rightarrow S=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P=|2x+1|+|-2y-4|\(\ge\left|2\left(x-y\right)-3\right|=3-2\sqrt{2}\)
=> a=3, b=-2
=> S=-1
Dấu bằng xảy ra khi x\(\ge\frac{-1}{2}\), y\(\le\)-2
Vậy S=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=2\sqrt{2\left(a^2+ab+2bc+2ca\right)}\)
\(=2\sqrt{2\left(a+b\right)\left(a+2c\right)}\le2\left(a+b\right)+\left(a+2c\right)=3a+2b+2c\)
Tương tự \(\sqrt{8b^2+56}\le2a+3b+2c;\)\(\sqrt{4c^2+7}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\)
Do vậy \(Q\ge\frac{11a+11b+12c}{3a+2b+2c+2a+3b+2c+\frac{a+b+4c}{2}}=2\)
Dấu "=" xảy ra khi và chỉ khi \(\left(a,b,c\right)=\left(1;1;\frac{3}{2}\right)\)
a) \(P=1957\)
b) \(S=19.\)