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\(4A=12x^2+12y^2+4z^2+20xy-12yz-12zx-8x-8y+12\)
\(=9x^2+9y^2+4z^2+18xy-12yz-12zx+2\left(x^2+y^2+4-4x-4y+2xy\right)+x^2+y^2-2xy+4\)
\(=\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\)
Dấu \(=\)khi \(\hept{\begin{cases}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\).
Vậy \(minA=1\)khi \(x=y=1,z=3\).
\(A=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}\left(x^2y^2+\frac{2}{3}xy-\frac{8}{3}x-\frac{8}{3}y\right)+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{4}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}y^2-\frac{16}{9}y-\frac{16}{9}]\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2-\frac{2y}{9}]+3\)
\(=\left(z-\frac{3}{2}x-\frac{3}{2}y\right)^2+\frac{3}{y}[\left(x+\frac{y}{3}-\frac{4}{3}\right)^2+\frac{8}{9}\left(y-1\right)^2]+1\)
\(\Leftrightarrow A\ge1\Leftrightarrow MinA=1\)
Dấu '' = '' xảy ra khi:
\(\hept{\begin{cases}z-\frac{3}{2}x-\frac{3}{2}y=0\\y-1=0\\x+\frac{y}{3}-\frac{4}{3}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}z=0\\y=1\\x=1\end{cases}}\)
A = x2 - 2xy + 3y2 - 2x + 1997
= ( x2 - 2xy + y2 - 2x + 2y + 1 ) + ( 2y2 - 2y + 1/2 ) + 3991/2
= [ ( x2 - 2xy + y2 ) - ( 2x - 2y ) + 1 ] + 2( y2 - y + 1/4 ) + 3991/2
= [ ( x - y )2 - 2( x - y ) + 12 ] + 2( y - 1/2 )2 + 3991/2
= ( x - y - 1 )2 + 2( y - 1/2 )2 + 3991/2 ≥ 3991/2 ∀ x, y
Dấu "=" xảy ra <=> x = 3/2 ; y = 1/2
=> MinA = 3991/2 <=> x = 3/2 ; y = 1/2
a) 5xy² . (-3y)²
= 5xy² . 9y²
= (5.9).x.(y².y²)
= 45xy⁴
Hệ số: 45
Bậc: 5
b) x²yz . (-2xy)³
= x²yz . (-8x³y³)
= -8.(x².x³).(y.y³).z
= -8x⁵y⁴z
Hệ số: -8
Bậc: 10
c) (-2x²y)².8x³yz³
= 4x⁴y².8x³yz³
= (4.8).(x⁴.x³).(y².y).z³
= 32x⁷y³z³
Hệ số: 32
Bậc: 13
d) (-2xy³)².(-2xyz)³
= 4x²y⁶.(-8x³y³z³)
= [4.(-8)].(x².x³).(y⁶.y³).z³
= -32x⁵y⁹z³
Hệ số: -32
Bậc: 17
e) (-5xy³z).(-4x²)²
= (-5xy³z).(16x⁴)
= (-5.16).(x.x⁴).y³.z
= -80x⁵y³z
Hệ số: -80
Bậc: 9
f) (2x²y³)².(-2xy)
= (4x⁴y⁶).(-2xy)
= [4.(-2)].(x⁴.x).(y⁶.y)
= -8x⁵y⁷
Hệ số: -8
Bậc: 12
a: =5xy^2*9y^2=45xy^4
b: =x^2yz*(-8)x^3y^3=-8x^5y^4z
c: =4x^4y^2*8x^3yz^3=32x^7y^3z^3
d: =4x^2y^6*(-8)x^3y^3z^3=-32x^5y^9z^3
e: =-5xy^3z*16x^4=-80x^5y^3z
f: =4x^4y^6*(-2xy)=-8x^5y^7
\(B=3x^2+3y^2+z^2+5xy-3yz-3xz-2x-2y+3\\\Rightarrow4A=12x^2+12y^2+4z^2+20xy-12yz-12xz-8x-8y+12\\\\=[(9x^2+18xy+9y^2)-(12xz+12yz)+4z^2]+[(2x^2+4xy+2y^2)-(8x+8y)+8]+(x^2-2xy+y^2)+4\\=[(3x+3y)^2-2\cdot(3x+3y)\cdot2z+(2z)^2]+[2(x^2+2xy+y^2)-8(x+y)+8]+(x-y)^2+4\\=(3x+3y-2z)^2+2[(x+y)^2-4(x+y)+4]+(x-y)^2+4\\=(3x+3y-2z)^2+2(x+y-2)^2+(x-y)^2+4\)
Ta thấy: \(\left\{{}\begin{matrix}\left(3x+3y-2z\right)^2\ge0\forall x,y,z\\2\left(x+y-2\right)^2\ge0\forall x,y\\\left(x-y\right)^2\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left(3x+3y-2z\right)^2+2\left(x+y-2\right)^2+\left(x-y\right)^2+4\ge4\forall x,y,z\)
\(\Leftrightarrow4B\ge4\Leftrightarrow B\ge1\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}3x+3y-2z=0\\x+y-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x\\2x=2\\2z=6x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=3\end{matrix}\right.\)
Vậy \(Min_B=1\) khi \(x=y=1;z=3\).
\(Toru\)