ko có biết

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

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 \(x=y=z=\frac{1}{3}\).

\(\Leftrightarrow\left(a.a+a.(−2)−1a−1.−2)(a−3)(a−1\right)\)

\(\Leftrightarrow\left(a^2-3a+2\right)\left(a-3\right)\left(a-1\right)\)

\(\Leftrightarrow\left(a^3-6a^2+11a-6\right)\left(a-1\right)\)

\(\Leftrightarrow a^4-7a^3+17a^2-17a+6\)

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

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

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

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

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

Vậy ...

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

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

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

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

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

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

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

\(\Leftrightarrow4x^2+12x-5x-15-\left(14x+4x^2-21-6x\right)=0\)

\(\Leftrightarrow4x^2+7x-15-\left(4x^2+8x-21\right)=0\)

\(\Leftrightarrow-x+6=0\)

\(\Leftrightarrow x=6\)

đẹp hong mình vẽ đó