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\(x^2+2xy+y^2+6\left(x+y\right)+8=-y^2\)
\(\Leftrightarrow\left(x+y\right)^2+6\left(x+y\right)+8\le0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y+4\right)\le0\)
\(\Rightarrow-4\le x+y\le-2\)
\(\Rightarrow2016\le B\le2018\)
\(B_{min}=2016\) khi \(\left(x;y\right)=\left(-4;0\right)\)
\(B_{max}=2018\) khi \(\left(x;y\right)=\left(-2;0\right)\)
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\(A=x-2y+3\Rightarrow x=A+2y-3\)
\(\Rightarrow\left(2y+A-3\right)^2+y\left(A+2y-3\right)+2y^2=1\)
\(\Leftrightarrow8y^2+\left(5A-15\right)y+A^2-6A+8=0\)
\(\Delta=\left(5A-15\right)^2-32\left(A^2-6A+8\right)\ge0\)
\(\Leftrightarrow-7A^2+42A-31\ge0\)
\(\Rightarrow\dfrac{21-4\sqrt{14}}{7}\le A\le\dfrac{21+4\sqrt{14}}{7}\)
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ta có:
\(x+2y=3\Leftrightarrow x=3-2y\)
thay vào P, ta có:
\(P=\left(3-2y\right)^2+5y^2\)
\(P=\left(3y-2\right)^2+5\)
\(\Rightarrow P\ge5\)(dấu xảy ra dấu "="\(\Leftrightarrow x=y=\frac{2}{3}\))
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