viết lại pt dưới dạng 

\(x^2-2x\left(y+2\right)+\left(2y^2+8\right)=0.\)

\(\Delta`x=\left(y+2\right)^2-\left(2y^2+8\right)=0\)

\(\Delta`=y^2+4y+4-2y^2-8=-y^2+4y-4=0\)

\(\Delta`=-\left(y-2\right)^2=0\Leftrightarrow y=2\)

thay y=2 

\(x^2-4x+8-4x=-8\)

\(x^2-8x+16=0\)

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

        \(x^2-2xy+2y^2-4x=-8\)

\(\Leftrightarrow x^2-2xy+2y^2-4x+8=0\)

\(\Leftrightarrow2x^2-4xy+4y^2-8x+16=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-8x+16\right)=0\)

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

Ta có: \(\left(x-2y\right)^2+\left(x-4\right)^2\ge0\) \(\forall x;y\)

Dấu "=" xảy ra: \(\Leftrightarrow\hept{\begin{cases}x-2y=0\\x-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2y\\x=4\end{cases}\Leftrightarrow\hept{\begin{cases}y=2\\x=4\end{cases}}}\) (thỏa mãn)

Vậy x = 4 và y = 2

Bài bạn gửi hay đấy .Chúc bạn học tốt.

- Help Me :((

\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)

\(25\left(x^2+6x+9\right)+1-25x^2=0\)

\(25x^2+150x+225+1-25x^2=0\)

\(150x=-226\)

\(x=-\frac{113}{75}\)

\(M=2\left(x^3-y^3\right)-3\left(x^2+y^2\right)\)

\(=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+y^2\right)\)

\(=2\left(x^2+xy+y^2\right)-3\left(x^2+y^2\right)\)   (do x+y = 1)

\(=2x^2+2xy+2y^2-3x^2-3y^2\)

\(-x^2+2xy-y^2=-\left(x-y\right)^2=-1\)

do x+y =1 la sao a

\(\left(x+y+z\right)^2\)

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

\(=x^2+2xy+y^2+2xz+2xy+z^2\)

\(C=x^2+x+\frac{1}{2}\)

\(C=\text{[}x^2+2.\frac{1}{2}.x+\left(\frac{1}{2}\right)^2\text{]}+\frac{1}{4}\)

\(C=\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\)

Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)

\(C_{min}=\frac{1}{4}\Leftrightarrow\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(C_{min}=\frac{1}{4}\Leftrightarrow x=-\frac{1}{2}\)

Tham khảo nhé~