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\(A=x^2+y^2+\left(\dfrac{1}{2}\right)^2-2xy+2.\dfrac{1}{2}x-2.\dfrac{1}{2}.y+\dfrac{3}{4}\)
\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)
\(a,=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)
Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)
\(b,=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(c,=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
Áp dụng Bunyakovsky, ta có :
\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)
=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)
=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)
Mấy cái kia tương tự
A = x 2 + 2 x y + y 2 – 4 x – 4 y + 1 = ( x 2 + 2 x y + y 2 ) – ( 4 x + 4 y ) + 1 = ( x + y ) 2 – 4 ( x + y ) + 1
Tại x + y = 3, ta có: A = 3 2 – 4.3 + 1 = -2
Đáp án cần chọn là: D
Lời giải:
\(x^2+3y^2+10x-14y-2xy=11\)
$\Leftrightarrow (x^2-2xy+y^2)+2y^2+10x-14y=11$
$\Leftrightarrow (x-y)^2+10(x-y)+25+(2y^2-4y+2)=38$
$\Leftrightarrow (x-y+5)^2+2(y-1)^2=38$
$\Rightarrow (x-y+5)^2=38-2(y-1)^2\leq 38$
$\Rightarrow -\sqrt{38}\leq x-y+5\leq \sqrt{38}$
$\Leftrightarrow -\sqrt{38}-5\leq x-y\leq \sqrt{38}-5$
Vậy $A_{\min}=-\sqrt{38}-5$ và $A_{\max}=\sqrt{38}-5$
\(P=\dfrac{x^2-6xy+6y^2}{x^2-2xy+y^2}=\dfrac{-3\left(x^2-2xy+y^2\right)+4x^2-12xy+9y^2}{x^2-2xy+y^2}\)
\(=-3+\left(\dfrac{2x-3y}{x-y}\right)^2\ge-3\)
\(P_{min}=-3\) khi \(2x=3y\)
\(1,Sửa:A=4x^4+4x^2y+y^2+2=\left(2x^2+y\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow2x^2+y=0\Leftrightarrow x^2=-\dfrac{y}{2}\\ 2,B=\left(x+y\right)^2+\left(y+1\right)^2+12\ge12\\ B_{min}=12\Leftrightarrow\left\{{}\begin{matrix}x=-y=1\\y=-1\end{matrix}\right.\)