\(2x^2+y^2+2xy-8x-6y+10=0\)

\(\Rightarrow2.\left(2x^2+y^2+2xy-8x-6y+10\right)=0\)

\(\Rightarrow4x^2+2y^2+4xy-16x-12y+20=0\)

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

\(\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2=0\left(1\right)\)

Ta có: \(\hept{\begin{cases}\left(2x+y-4\right)^2\ge0\forall x;y\\\left(y-2\right)^2\ge0\forall y\end{cases}\Rightarrow\left(2x+y-4\right)^2+\left(y-2\right)^2\ge0\forall x;y\left(2\right)}\)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}2x+y-4=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+y=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}2x+2=4\\y=2\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

Chúc bạn học tốt.

\(x^2-14x+13=0\)

\(\Rightarrow x^2-2x.7+7^2-7^2+13=0\)

\(\Rightarrow\left(x^2-2x.7+7^2\right)-7^2+13=0\)

\(\Rightarrow\left(x-7\right)^2-49+13=0\)

\(\Rightarrow\left(x-7\right)^2-36=0\)

\(\Rightarrow\left(x-7\right)^2=36\)

\(\Rightarrow\left(x-7\right)^2=\pm6^2\)

\(\Rightarrow\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

Vậy ...

\(x^2-14x+13=0\)

\(x^2-14x+49-36=0\)

\(\left(x^2-14x+19\right)-36=0\)

\(\left(x-7\right)^2-6^2=0\)

\(\left(x-7-6\right)\left(x-7+6\right)=0\)

\(\left(x-13\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-13=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

\(9x^2+24xy+16y^2\)

\(=\left(3x\right)^2+2\cdot3x\cdot4y+\left(4y\right)^2\)

\(=\left(3x+4y\right)^2\)

\(8x^3+1=\left(2x\right)^3+1^3\)

\(=\left(2x+1\right)\left(4x^2+2x+1\right)\)

\(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)

\(\left(a^2+9\right)^2-36a^2\)

\(=\left(a^2+9-6a\right)\left(a^2+9+6a\right)\)

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

Ta có:

\(\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right)\div\frac{2-4x}{x+1}\)

\(=\frac{\left(x+2\right)\left(x+1\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}\cdot\frac{x+1}{2-4x}\)

\(=\frac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\frac{1}{2\left(1-2x\right)}\)

\(=\frac{-8x^2+2}{3x}\cdot\frac{1}{2\left(1-2x\right)}\)

\(=\frac{-2\left(4x^2-1\right)}{3x}\cdot\frac{-1}{2\left(2x-1\right)}\)

\(=\frac{-2\left(2x+1\right)\left(2x-1\right)}{3x}\cdot\frac{-1}{2\left(2x-1\right)}\)

\(=\frac{2x+1}{3x}\)