\(Q=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Ta có: 

\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)

Tương tự: 

\(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)

\(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)

Cộng lại vế với vế ta được: 

\(\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y-z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}=0\)

Suy ra \(2Q=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\ge\frac{\frac{\left(x^2+y^2\right)^2}{2}}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{\frac{\left(z^2+x^2\right)^2}{2}}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)

\(\ge\frac{\frac{\left(x+y\right)^2}{2}}{2\left(x+y\right)}+\frac{\frac{\left(y+z\right)^2}{2}}{2\left(y+z\right)}+\frac{\frac{\left(z+x\right)^2}{2}}{2\left(z+x\right)}\)

\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}\)

\(=\frac{x+y+z}{2}=\frac{1}{2}\)

Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{3}\).

Vậy \(minQ=\frac{1}{4}\)đạt tại \(x=y=z=\frac{1}{3}\).

\(\left|3-x\right|+x^2-\left(4+x\right)x=0\)

\(\left|3-x\right|-4x=0\)

\(\left|3-x\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=4x\\3-x=-4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=3\\-3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-1\end{matrix}\right.\)

Vậy ...

\(\Leftrightarrow\left|3-x\right|+x^2-4x-x^2=0\)

\(\Leftrightarrow\left|3-x\right|-4x=0\)

\(\Leftrightarrow\left|3-x\right|=4x\)

\(\Leftrightarrow\orbr{\begin{cases}3-x=4x\\3-x=-4x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-5x=-3\\3x=-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-1\end{cases}}\)

\(S=\left\{\frac{3}{5};-1\right\}\)

đẹp hong mình vẽ đó

`Answer:`

Giải hệ phương trình à bạn?

\(\hept{\begin{cases}56x+65y=35,4\\x+y=0,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}56x+65y=35,4\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}9y=1,8\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x+56y=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x+56.0,2=33,6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\56x=33,6-11,2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=0,2\\x=0,4\end{cases}}\)

Cuộc chiến chống Pháp: Phan Đình Giót, Tô Vĩnh Diện, Bế Văn Đàn, Trần Can.

em bik từng đấy thôi chị nha

wow đúng rồi

\(\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)\ge\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)^2\)

\(\Rightarrow\sqrt{a^2+\dfrac{1}{b+c}}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)=\dfrac{1}{\sqrt{17}}\left(4a+\dfrac{1}{\sqrt{b+c}}\right)\)

Tương tự:

\(\sqrt{b^2+\dfrac{1}{a+c}}\ge\dfrac{1}{\sqrt{17}}\left(4b+\dfrac{1}{\sqrt{a+c}}\right)\) ; \(\sqrt{c^2+\dfrac{1}{a+b}}\ge\dfrac{1}{\sqrt{17}}\left(4c+\dfrac{1}{\sqrt{a+b}}\right)\)

Cộng vế:

\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)

\(VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)

Cũng theo Bunhiacopxki:

\(1.\sqrt{a+b}+1.\sqrt{b+c}+1\sqrt{c+a}\le\sqrt{\left(1+1+1\right)\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)

\(\Rightarrow VT\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)

\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\) 

\(VT\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=2\)

Bài này giải bằng Bunhiacopxki (kết hợp nguyên lý Dirichlet) chứ AM-GM thì e là không ổn:

Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có 2 số cùng phía so với 1, không mất tính tổng quát, giả sử đó là \(b^2\) và \(c^2\)

\(\Rightarrow\left(b^2-1\right)\left(c^2-1\right)\ge0\)

\(\Rightarrow b^2c^2+1\ge b^2+c^2\)

\(\Rightarrow b^2c^2+2b^2+2c^2+4\ge3b^2+3c^2+3\)

\(\Rightarrow\left(b^2+2\right)\left(c^2+2\right)\ge3\left(b^2+c^2+1\right)\)

\(\Rightarrow\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\ge3\left(a^2+1+1\right)\left(1+b^2+c^2\right)\ge3\left(a+b+c\right)^2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\left(2+7\right)\left(2a^2+\dfrac{7}{b^2}\right)\ge\left(2a+\dfrac{7}{b}\right)^2\)

\(\Rightarrow\sqrt{2a^2+\dfrac{7}{b^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{b}\right)\)

Tương tự: \(\sqrt{2b^2+\dfrac{7}{c^2}}\ge\dfrac{1}{3}\left(2a+\dfrac{7}{c}\right)\) ; \(\sqrt{2c^2+\dfrac{7}{a^2}}\ge\dfrac{1}{3}\left(2c+\dfrac{7}{a}\right)\)

Cộng vế:

\(VT\ge\dfrac{1}{3}\left(2a+2b+2c+\dfrac{7}{a}+\dfrac{7}{b}+\dfrac{7}{c}\right)=2+\dfrac{7}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(VT\ge2+\dfrac{7}{9}.\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) (do \(a+b+c=3\))

\(VT\ge2+\dfrac{7}{9}.\left(\sqrt{a}.\sqrt{\dfrac{1}{a}}+\sqrt{b}.\sqrt{\dfrac{1}{b}}+\sqrt{c}.\sqrt{\dfrac{1}{c}}\right)^2=9\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)