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\(2M=4x^2+10y^2-4xy+4x+4y\)
\(2M=4x^2+y^2+1-4xy+4x-2y+9y^2+6y+1-2\)
\(2M=\left(2x-y+1\right)^2+\left(3y+1\right)^2-2\ge-2\)
\(\Rightarrow M\ge-1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y=-\frac{1}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
a/ A = 2x2 + y2 - 2xy - 2x + 3
= (x2 - 2xy + y2) + (x2 - 2x + 1) + 2
= (x - y)2 + (x - 1)2 + 2\(\ge2\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
\(D=\frac{1}{2}\left(4x^2+4xy+y^2+16-16x-8y\right)+\frac{9}{2}\left(y^2-4y+4\right)-26\)
\(D=\frac{1}{2}\left(2x+y-4\right)^2+\frac{9}{2}\left(y-2\right)^2-26\ge-26\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)
\(2x^2+y^2+2xy-6x-2y+10\)
\(=\left(x^2-4x+4\right)+\left(x^2+y^2+1+2xy-2y-2x\right)+5\)
\(=\left(x-2\right)^2+\left(x+y-1\right)^2+5\ge5\)
a)\(2x^2+y^2+4x-2y-2xy+10=2x^2+y^2+4x-2y\left(x+1\right)+10\)
\(=y^2-2y\left(x+1\right)+2\left(x^2+2x+1\right)+8\)
\(=y^2-2y\left(x+1\right)+2\left(x+1\right)^2+8\)
\(=\left(y+x+1\right)^2+\left(x+1\right)^2+8\ge8\)
Dấu "=" xảy ra khi x=-1 và y=0
\(M=2x^2+5y^2-2xy+2y+2x=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(4y^2+2y+\dfrac{1}{4}\right)-\dfrac{5}{4}=\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\dfrac{1}{2}\right)^2-\dfrac{5}{4}\)ta có: (x - y)^2 ≥ 0; (x+1)^2≥ ; (2y+1/2)^2 ≥ 0
=> gtnn M = -5/4
ách nhầm:
\(M=2x^2+5y^2-2xy+2y+2x=\left(x^2+2x+1\right)+\left(x^2-2xy+y^2\right)+4\left(y^2+\dfrac{1}{2}y+\dfrac{1}{16}\right)+\dfrac{3}{4}=\left(x+1\right)^2+\left(x-y\right)^2+4\left(y-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\)
ta có: (x - y)^2 ≥ 0; (x+1)^2≥ ; 4(y+1/4)^2 ≥ 0
vậy gtnn M = 3/4 khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x+1\right)^2=0\\\left(y-\dfrac{1}{4}\right)^2=0\end{matrix}\right.\)
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