Phòng GD-ĐT TP. Bắc Giang năm học 2015-2016

Bài 2:

a/  M=2x²+5y²-6xy+4x-10y+100
<=>M= 1/2(4x²+10y²-12xy+8x-20y+200)

<=>M=1/2[(4x²+9y²+4-12xy+8x-12y)+(y²-8y+16)+180]

<=>M=1/2[(2x-3y+2)²+(y-4)²+180]

<=>M=1/2(2x-3y+2)²+1/2(y-4)²+90

1/2(2x-3y+2)²+1/2(y-4)² >=0

=> M >= 90
Dấu "=" xảy ra <=> (x,y)=(5;4)

Vậy min M là M=90 tại (x,y)=(5;4)

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a) \(x^2+5y^2+2x-4xy-10y+14\)

\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+\left(y^2-6y+9\right)+4\)

\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y-3\right)^2+4\)

\(=\left(x-2y+1\right)^2+\left(y-3\right)^2+4\)

Vì \(\left(x-2y+1\right)^2\ge0\)

      \(\left(y-3\right)^2\ge0\)

 \(\Rightarrow\left(x-2y+1\right)^2+\left(y-3\right)^2+4\ge4>0\)với mọi x,y (ĐPCM)
b) \(5x^2+10y^2-6xy-4x-2y+3\)

\(=\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^2\right)+\left(y^2-2y+1\right)+1\)

\(=\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-1\right)^2+1\)

Vì \(\left(2x-1\right)^2\ge0\)

      \(\left(x-3y\right)^2\ge0\)

       \(\left(y-1\right)^2\ge0\)

 \(\Rightarrow\left(2x-1\right)^2+\left(x-3y\right)^2+\left(y-1\right)^2+1\ge1>0\)vợi mọi x,y (ĐPCM)

Nỗi hứng lm cho vui!

Bài 1:

a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)

Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12

=> Dấu = xảy ra <=> \(x=2\)

b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)

= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)

= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)

= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)

Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)

Bài 2:

a) P = \(-x^2-4x+16=-\left(x^2+4x+4\right)+20\)

= \(-\left(x+2\right)^2+20\le20\)

=> Dấu = xảy ra <=> \(x=-2\)

b) \(Q=-x^2+2xy-4y^2+2x+10y-2017\)

= \(-\left[\left(x^2-2xy+y^2\right)+3\left(y^2-4y+4\right)-2\left(x-y\right)+2005\right]\)

= \(-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2004\right]\)

= \(-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]-2004\)

Vì \(\left\{{}\begin{matrix}-\left(x-y-1\right)^2\le0\\3\left(y-2\right)^2\le0\end{matrix}\right.\) => Q \(\le-2004\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)

đăng từng câu 1 thôi

Ta có: \(C=10x^2+4x-12xy+5y^2+6y+19\)

\(=\left(y^2+6y+9\right)+\left(4y^2-12xy+9x^2\right)+\left(x^2+4x+4\right)+6\)

\(=\left(y+3\right)^2+\left(2y-3x\right)^2+\left(x+2\right)^2+6\ge6\forall x,y\)

Dấu '=' xảy ra khi:

\(\left\{{}\begin{matrix}y+3=0\\2y-3x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=-2\end{matrix}\right.\)

Vậy: Giá trị nhỏ nhất của biểu thức \(C=10x^2+4x-12xy+5y^2+6y+19\) là 6 khi x=-2 và y=-3