tìm x y thỏa mãn
3*(2x-1)^2 + 7*(3y+5 )^2 = 0
[tex]\dpi{100} (2x-1)^2\geq 0\forall x\Rightarrow 3.(2x-1)^2\geq 0\forall x(1)\\(3y+5)^2\geq 0\forall y\Rightarrow 7.(3y+5)^2\geq 0\forall y(2)\\(1),(2)\Rightarrow 3.(2x-1)^2+7.(3y+5)^2\geq 0\forall x,y(3)\\De,(3)\Rightarrow 3.(2x-1)^2+7.(3y+5)^2=0\\\Rightarrow \left\{\begin{matrix} 3.(2x-1)^2=0\\ 7.(3y+5)^2=0 \end{matrix}\right.\\\Leftrightarrow \left\{\begin{matrix} (2x-1)^2=0\\ (3y+5)^2=0 \end{matrix}\right.\\\Leftrightarrow \left\{\begin{matrix} 2x-1=0\\ 3y+5=0 \end{matrix}\right.\\\Leftrightarrow \left\{\begin{matrix} 2x=1\\ 3y=-5 \end{matrix}\right.\\\Leftrightarrow \left\{\begin{matrix} x=\frac{1}{2}\\ y=\frac{-5}{3} \end{matrix}\right.[/tex]
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