\(\frac{a}{b}=\frac{c}{d}\Rightarrow a.d=b.c\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{5a}{5c}=\frac{5b}{5d}=\frac{5a+5b}{5c+5d}=\frac{5a-5b}{5c-5d}\)

\(\Rightarrow\frac{5a+5b}{5c+5d}=\frac{5a-5b}{5c-5d}\)

      \(5x^2+9y^2-12xy-6x+9=0\)

\(\Rightarrow\left(4x^2+9y^2-12xy\right)+\left(x^2-6x+9\right)=0\)

\(\Rightarrow\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3y\\x=3\end{cases}\Rightarrow}\hept{\begin{cases}y=2\\x=3\end{cases}}}\)

\(5x^2+9y^2-12xy-6x+9=0\)

<=>  \(\left(4x^2-12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)

<=>  \(\left(2x-3y\right)^2+\left(x-3\right)^2=0\)

<=>  \(\hept{\begin{cases}2x-3y=0\\x-3=0\end{cases}}\)

<=>  \(\hept{\begin{cases}y=2\\x=3\end{cases}}\)

Vậy...

\(x^2+y^2-2x+4y+5=0\)

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

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

Ta thấy: \(\left(x-1\right)^2\ge0;\left(y+2\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\)

Do đó: \(\hept{\begin{cases}x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}x=1\\y=-2\end{cases}}\). Vậy ...

u15=u13+u14

u20=u18+u19