\(4x^2+8x+2=0\)\(\Leftrightarrow4x^2+8x+4-2=0\)\(\Leftrightarrow4\left(x^2+2x+1\right)-2=0\)\(\Leftrightarrow4\left(x+1\right)^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)\right]^2-\left(\sqrt{2}\right)^2=0\)\(\Leftrightarrow\left[2\left(x+1\right)+\sqrt{2}\right]\left[2\left(x+1\right)-\sqrt{2}\right]=0\)\(\Leftrightarrow\left(2x+2+\sqrt{2}\right)\left(2x+2-\sqrt{2}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x+2+\sqrt{2}=0\\2x+2-\sqrt{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2-\sqrt{2}}{2}\\x=\frac{-2+\sqrt{2}}{2}\end{cases}}\)

Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{-2\pm\sqrt{2}}{2}\right\}\)

ơ ơ ko lm đc 

thay 2 => 4 may ra còn lm đc

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

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.

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