Bài 2:a. 2x^2+2xy+y^2+96x-left|y+3right| Leftrightarrowleft|y+3right|6x-2x^2-2xy-y^2-9 Leftrightarrowleft|y+3right|-x^2-2xy-y^2-x^2+6x-9 Leftrightarrowleft|y+3right|-left(x+yright)^2-left(x-3right)^2 Leftrightarrowleft|y+3right|-left[left(x+yright)^2+left(x-3right)^2right] Có: left|y+3right|ge0 -left[left(x+yright)^2+left(x-3right)^2right]le0 Do đó: left|y+3right|-left[left(x+yright)^2+left(x-3right)^2right]0 Leftrightarrowhept{begin{cases}y+30x+y0x-30end{cases}}Leftrightarrowhept{begin{cases}x3...
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Bài 2:
a. \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow\left|y+3\right|=6x-2x^2-2xy-y^2-9\)
\(\Leftrightarrow\left|y+3\right|=-x^2-2xy-y^2-x^2+6x-9\)
\(\Leftrightarrow\left|y+3\right|=-\left(x+y\right)^2-\left(x-3\right)^2\)
\(\Leftrightarrow\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\)
Có: \(\left|y+3\right|\ge0\)
\(-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\le0\)
Do đó: \(\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}y+3=0\\x+y=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
b. \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+\left[2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\left(2x^2+x-2013-2x^2+10x+4024\right)^2=0\)
\(\Leftrightarrow\left(11x+2011\right)^2=0\)
\(\Leftrightarrow11x+2011=0\)
\(\Leftrightarrow x=-\frac{2011}{11}\)
