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bài 1 : a, sai đề . đề đúng là x3 + x2 + 22 = x3 + 2x2- x2+ 4
= x2 ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 ) ( x2 - x + 2 )
b, xy -y2 -x -y = ( xy -x ) + ( y2 - y ) = x ( y -1) + y ( y -1) = ( y -1 ) ( x +y )
a )x2+2y2-2xy+2x-4y+2=0
<=>x2-2x(y-1)+y2-2y+1+y2-2y+1=0
<=>x2-2x(y-1)+(y-1)2+(y-1)2=0
<=>(x-y+1)2+(y-1)2=0
<=>x-y+1=0 va y-1=0
<=>x=y-1 y=1
<=>x=1-1=0 y=1
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
\(x^2+4y^2-2x+4y+2=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
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