Bài 5.`@`\(A=x^2-2x+5\)\(A=\left(x^2-2x+1\right)+4\)\(A=\left(x-1\right)^2+4\ge4\)Dấu "=" xảy ra ` x-1=0` ` x=1`Vậy \(Min_A=4\) khi `x=1``@`\(B=x^2+4y^2+2x-4y+5\)\(B=\left(x^2+2x+1\right)+\left(4y^2-4y+1\right)+3\)\(B=\left(x+1\right)^2+\left(2y-1\right)^2+3\ge3\)Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{2}\end{matrix}\right.\)Vậy \(Min_B=3\) khi \(\left(x;y\right)=\left(-1;\dfrac{1}{2}\right)\)`@`\(C=5x^2+y^2+4xy-8x-2y+10\)\(C=\left[y^2+2y\left(2x-1\right)+\left(2x-1\right)^2\right]+\left(x^2-4x+4\right)+5\)\(C=\left(y+2x-1\right)^2+\left(x-2\right)^2+5\ge5\)Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y+2x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)Vậy \(Min_C=5\) khi \(\left(x;y\right)=\left(2;-3\right)\) Câu trả lời: Bài 5.`@`\(A=x^2-2x+5\)\(A=\left(x^2-2x+1\right)+4\)\(A=\left(x-1\right)^2+4\ge4\)Dấu "=" xảy ra ` x-1=0` ` x=1`Vậy \(Min_A=4\) khi `x=1``@`\(B=x^2+4y^2+2x-4y+5\)\(B=\left(x^2+2x+1\right)+\left(4y^2-4y+1\right)+3\)\(B=\left(x+1\right)^2+\left(2y-1\right)^2+3\ge3\)Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{2}\end{matrix}\right.\)Vậy \(Min_B=3\) khi \(\left(x;y\right)=\left(-1;\dfrac{1}{2}\right)\)`@`\(C=5x^2+y^2+4xy-8x-2y+10\)\(C=\left[y^2+2y\left(2x-1\right)+\left(2x-1\right)^2\right]+\left(x^2-4x+4\right)+5\)\(C=\left(y+2x-1\right)^2+\left(x-2\right)^2+5\ge5\)Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y+2x-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)Vậy \(Min_C=5\) khi \(\left(x;y\right)=\left(2;-3\right)\)