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\(\text{a) }2x^2+y^2-2xy-2x+3\\ =\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\\ =\left(x-y\right)^2+\left(x-1\right)^2+2\)
\(\text{Ta có : }\left(x-y\right)^2\ge0\\ \left(x-1\right)^2\ge0\\ \Rightarrow\left(x-y\right)^2+\left(x-1\right)^2\ge0\\ \Rightarrow\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\) \(\text{Dấu }"="\text{xảy ra khi : }\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1-y=0\\x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Vậy giá trị nhỏ nhất biểu thức \(=2\) khi \(x=1\) và \(y=1\)
\(\text{b) }x\left(x-1\right)\left(x+4\right)\left(x+3\right)+15\left(Chữa\text{ }đề\text{ }nhé\right)\\ =\left(x^2+3x\right)\left(x^2-x+4x-4\right)+15\\ =\left(x^2+3x\right)\left(x^2+3x-4\right)+15\left(1\right)\\ Đặt\text{ }x^2+3x=y\left(2\right)\\ Thay\text{ }\left(2\right)\text{ }vào\text{ }\left(1\right),ta\text{ }được:\text{ }\left(1\right)=y\left(y-4\right)+15\\ =y^2-4y+15\\ =\left(y^2-4y+4\right)+11\\ =\left(y-2\right)^2+11\\ Do\text{ }\left(y-2\right)^2\ge0\forall x\\ \Rightarrow\left(y-2\right)^2+11\ge11\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\left(y-2\right)^2=0\\\Leftrightarrow y-2=0\\\Leftrightarrow x^2+3x-2=0\\\Leftrightarrow x\left(x+3\right)=2 \)
Vậy GTNN của biểu thức là \(11\) khi \(x\left(x+3\right)=2\)
\(\text{c) }x^2-xy+y^2-2x-2y\\ \Rightarrow2S=2x^2-2xy+2y^2-4x-4y\\ =\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\\ =\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\\ Do\text{ }\left(x-y\right)^2\ge0\forall x;y\\ \left(x-2\right)^2\ge0\forall x\\ \left(y-2\right)^2\ge0\forall y\\ \Rightarrow\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\forall x;y\\ \Rightarrow2S=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\forall x;y\\ \Rightarrow S\ge-4\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-2=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=2\\y=2\end{matrix}\right.\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }-4\text{ }khi\text{ }x=y=2\)
