Tìm x,y biết: (a) 10x^2+37x+7=0. (b) 4x^2+9y^2+24y+25=12x.

a) \[10{x^2} + 36x + 7 = 0\]

\[10{x^2} + 2x + 35x + 7 = 0\]

\[\left( {10{x^2} + 2x} \right) + \left( {35x + 7} \right) = 0\]

\[2x\left( {5x + 1} \right) + 5\left( {5x + 1} \right) = 0\]

\[\begin{array}{l}\left( {5x + 1} \right)\left( {2x + 5} \right) = 0\\TH1:5x + 1 = 0 \Rightarrow x = \frac{{ - 1}}{5}\\TH2:2x + 5 = 0 \Rightarrow x = \frac{{ - 5}}{2}\end{array}\]

Vậy \(x \in \left\{ {\frac{{ - 1}}{5};\frac{{ - 5}}{2}} \right\}\).

b) \(\begin{array}{l}4{x^2} + 9{y^2} + 24y + 25 = 12x\\\left( {4{x^2} - 12x + 9} \right) + \left( {9{y^2} + 24y + 16} \right) = 0\\{\left( {2x - 3} \right)^2} + {\left( {3y + 4} \right)^2} = 0\\ \Rightarrow \left\{ \begin{array}{l}2x - 3 = 0\\3y + 4 = 0\end{array} \right.\\\left\{ \begin{array}{l}x = \frac{3}{2}\\y = \frac{{ - 4}}{3}\end{array} \right.\end{array}\)

Vậy \(x = \frac{3}{2};y = \frac{{ - 4}}{3}\).