\(E=\sqrt{\left(x-2016\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(=\left|x-2016\right|+\left|x-1\right|\)

\(=\left|x-2016\right|+\left|1-x\right|\ge\left|\left(x-2016\right)+\left(1-x\right)\right|=2015\)

(Dấu "="\(\Leftrightarrow\left(x-2016\right)\left(1-x\right)\ge0\)

\(TH1:\hept{\begin{cases}x-2016\ge0\\1-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2016\\x\le1\end{cases}}\left(L\right)\)

\(TH2:\hept{\begin{cases}x-2016\le0\\1-x\le0\end{cases}}\Leftrightarrow1\le x\le2016\))

Vậy \(E_{min}=2015\Leftrightarrow1\le x\le2016\)

Áp dụng BĐT |a|+|b|\(\ge\)|a+b| ta có:

\(E=\sqrt{\left(x-2016\right)^2}+\sqrt{\left(x-1\right)^2}\)

\(=\left|x-2016\right|+\left|x-1\right|\)

\(=\left|x-2016\right|+\left|-\left(x-1\right)\right|\)

\(=\left|x-2016\right|+\left|-x+1\right|\)

\(\ge\left|x-2016+\left(-x\right)+1\right|=2015\)

Xảy ra khi \(1\le x\le2016\)

Nguyễn Huyền Anh

chtt đi bạn

min mình có ra rồi. nhưng chỉ không biết là khi x=y hay x,y bằng bao nhiêu thôi. 

\(\sqrt{x\left(x+3y\right)}\ge\frac{x+x+3y}{2}=\frac{2x+3y}{2}\)

\(\sqrt{y\left(y+3x\right)}\le\frac{y+y+3x}{2}=\frac{2y+3x}{2}\)

\(\Rightarrow\sqrt{x\left(x+3y\right)}+\sqrt{y\left(y+3x\right)}\le\frac{5}{2}\left(x+y\right)\)

=> \(A\ge2016\left(x+y\right):\frac{5}{2}\left(x+y\right)=\frac{2016\cdot2\left(x+y\right)}{5\left(x+y\right)}=\frac{4032}{5}\)

nhưng không biết x,y bằng bao nhiêu. ai làm đc ghi hẳn cách giải ra nha

... mẫu \(\le\sqrt{\left(\sqrt{x}^2+\sqrt{y}^2\right)\left(\sqrt{x+3y}^2+\sqrt{y+3x}^2\right)}\)=2(x+y)

=>A\(\ge\)(2016(x+y))/(2(x+y) =1008

=> Min A = 1008. Dấu x xảy ra <=> x=y

\(a)\) \(M_{\left(3\right)}=3+3^2+3^3+...+3^{2016}\)

\(3M_{\left(3\right)}=3^2+3^3+3^4+...+3^{2017}\)

\(3M_{\left(3\right)}-M_{\left(3\right)}=\left(3^2+3^3+3^4+...+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2016}\right)\)

\(2M_{\left(3\right)}=3^{2017}-3\)

\(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)

Vậy \(M_{\left(3\right)}=\frac{3^{2017}-3}{2}\)

\(M_{\left(-3\right)}=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2016}\)

\(\left(-3\right)M_{\left(-3\right)}=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+...+\left(-3\right)^{2017}\)

\(\left(-3\right)M_{\left(-3\right)}-M_{\left(-3\right)}=\left[\left(-3\right)^2+\left(-3\right)^3+...+\left(-3\right)^{2017}\right]-\left[\left(-3\right)+\left(-3\right)^2+...+\left(-3\right)^{2016}\right]\)\(\left(-4\right)M_{\left(-3\right)}=\left(-3\right)^{2017}+3\)

\(M_{\left(-3\right)}=\frac{\left(-3\right)^{2017}+3}{-4}\)

\(M_{\left(-3\right)}=\frac{-\left(3^{2017}-3\right)}{-4}\)

\(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)

Vậy \(M_{\left(-3\right)}=\frac{3^{2017}-3}{4}\)

Chúc bạn học tốt ~ 

\(b)\) Ta có : 

\(M_{\left(2\right)}=2+2^2+2^3+...+2^{2016}\)

\(M_{\left(2\right)}=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)

\(M_{\left(2\right)}=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)\)

\(M_{\left(2\right)}=2.7+2^4.7+...+2^{2014}.7\)

\(M_{\left(2\right)}=7\left(2+2^4+...+2^{2014}\right)⋮7\) \(\left(1\right)\)

Lại có : 

\(M_{\left(2\right)}=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{2013}+2^{2014}+2^{2015}+2^{2016}\right)\)

\(M_{\left(2\right)}=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{2013}\left(1+2+2^2+2^3\right)\)

\(M_{\left(2\right)}=2.15+2^5.15+...+2^{2013}.15\)

\(M_{\left(2\right)}=15\left(2+2^5+...+2^{2013}\right)⋮15\) \(\left(2\right)\)

Từ (1) và (2) suy ra \(M_{\left(2\right)}\) chia hết cho \(7\) và \(15\)

\(\Rightarrow\)\(M_{\left(2\right)}⋮105\) ( vì \(7.15=105\) ) 

Vậy nếu \(M⋮105\)\(\Leftrightarrow\)\(x=2\)

Chúc bạn học tốt ~ 

Áp dụng BĐT bunyakovsky:

\(\sum\dfrac{x^2}{y+z}\ge\sum\dfrac{x^2}{\sqrt{2\left(y^2+z^2\right)}}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+y^2}=a\\\sqrt{y^2+z^2}=b\\\sqrt{z^2+x^2}=c\end{matrix}\right.\) thì có a+b+c=2016 và cần tìm Min của \(\sum\dfrac{a^2+c^2-b^2}{2\sqrt{2}b}\) (\(x^2=\dfrac{a^2+c^2-b^2}{2}\))

Ta có: \(\sum\dfrac{a^2+c^2-b^2}{2\sqrt{2}b}=\dfrac{1}{2\sqrt{2}}.\left(\sum_{sym}\dfrac{a^2}{b}-\sum b\right)\)

Áp dụng BĐT cauchy-schwarz:

\(\sum_{sym}\dfrac{a^2}{b}=\dfrac{a^2}{b}+\dfrac{c^2}{b}+\dfrac{b^2}{a}+\dfrac{c^2}{a}+\dfrac{a^2}{c}+\dfrac{b^2}{c}\ge\dfrac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)}=2\left(a+b+c\right)\)

DO đó \(VT\ge\dfrac{1}{2\sqrt{2}}\left(2\sum a-\sum a\right)=\dfrac{1}{2\sqrt{2}}\left(a+b+c\right)=\dfrac{2016}{2\sqrt{2}}=\dfrac{1008}{\sqrt{2}}\)

Dấu = xảy ra khi a=b=c hay \(x=y=z=\dfrac{672}{\sqrt{2}}\)