2x2+9y2-6xy-6x-12y+2016
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A=2x^2+9y^2-6xy-6x-12y+2024
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2024
A = (x -3y)^2 +4(x -3y) + 4 + x^2 -10x +25 + 1995
A = (x -3y +2)^2 + (x -5)^2 + 1995 \geq 1995
Min A = 1995
x - 5 = 0 => x = 5
Và x - 3y + 2 = 0 hay 5 -3y +2 = 0 => -3y = -7 => y = 7/3
\(K\)\(nha!~!\)
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Bài làm
a) xy + y2 - x - y
= ( xy + y2 ) - ( x + y )
= y( x + y ) - ( x + y )
= ( x + y )( y - 1 )
b) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 25 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
c) xy + xz - 2y - 2z
= ( xy + xz ) - ( 2y + 2z )
= x( y + z ) - 2( y + z )
= ( y + z )( x - 2 )
d) x2 - 6xy + 9y2 - 25z2
= ( x2 - 6xy + 9y2 ) - 25z2
= ( x - 3y )2 - 25z2
= ( x - 3y - 5z )( z - 3y + 5z )
e) 3x2 - 3y2 - 12x + 12y
= 3( x - y )( x + y ) - 12( x - y )
= ( x - y )[ 3( x + y ) - 12 ]
f) 4x3 + 4xy2 + 8x2y - 16x
= 4x( x2 + y2 + 2xy - 4 )
= 4x[ ( x + y)2 - 4 ]
= 4x( x + y - 2 )( x + y + 2 )
g) x2 - 5x + 4
= x2 - x - 4x + 4
= x( x - 1 ) - 4( x - 1 )
= ( x - 1 )( x - 4 )
h) x4 + 5x2 + 4
= x4 + x2 + 4x2 + 4
= x2( x2 + 1 ) + 4( x2 + 1 )
= ( x2 + 1 )( x2 + 4 )
i) 2x2 + 3x - 5
= 2x2 - 5x + 2x - 5
= 2x( x + 1 ) - 5( x + 1 )
= ( x + 1 )( 2x - 5 )
k) x3 - 2x2 + 6x - 5 ( không biết làm )
l) x2 - 4x + 3
= ( x2 - 4x + 4 ) - 1
= ( x - 2 )2 - 1
= ( x - 3 )( x - 1 )
# Học tốt #
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\(B=2x^2+9y^2-6xy-6x-12y+2016\)
\(=x^2-6xy+9y^2+4x-12y+4+x^2-10x+25+1987\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-5\right)^2+1987\)
\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1987\)
Vì \(\left(x-3y+2\right)^2\ge0\) và \(\left(x-5\right)^2\ge0\) nên GTNN của B là 1987, đạt được khi
\(\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Leftrightarrow\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}\)
\(2x^2+9y^2-6xy-6x-12y+2008\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+4+\left(x^2-10x+25\right)+1979\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-5\right)^2+1979\)
\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+1979\ge1979\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\frac{7}{3}\end{matrix}\right.\)
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\(A=9y^2-6xy-12y+2x^2-6x+2016\)
\(A=\left(3y\right)^2-2.3y\left(x+2\right)+\left(x+2\right)^2-\left(x^2+4x+4\right)+2x^2-6x+2016\)
\(A=\left(3y-\left(x+2\right)\right)^2+x^2-10x+2012\)
\(A=\left(3y-x-2\right)^2+\left(x-5\right)^2+1987\)
Vậy GTNN của A = 1987 khi x=5 và y=7/3.
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a)\(x^4+3x^3+x^2+3x=x\left(x^3+3x^2+x+3\right)\)
\(=x\left[x^2\left(x+3\right)+\left(x+3\right)\right]=x\left(x+3\right)\left(x^2+1\right)\)
b) \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-4z^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)
c) \(=2x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(2x-7\right)\)
\(a,=x^3\left(x+3\right)+x\left(x+3\right)=x\left(x^2+1\right)\left(x+3\right)\\ b,=\left(x+3y\right)^2-4z^2=\left(x+3y+2z\right)\left(x+3y-2z\right)\\ c,=2x^2-2x-7x+7=\left(x-1\right)\left(2x-7\right)\)
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\(\Leftrightarrow\left(x^2-6xy+9y^2\right)+\left(x^2+6x+9\right)+\left(z^2-8z+16\right)=0\)
\(\Leftrightarrow\left(x-3y\right)^2+\left(x+3\right)^2+\left(z-4\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\x+3=0\\z-4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\\z=4\end{matrix}\right.\)
Sao ko cóp phần sau z
tìm min thì tách thích hợp ra toàn phương