Bài 1. a) E = x2 - 2x + y2 + 4y + 8
E = ( x2 - 2x + 1) + ( y2 + 2.2x + 22) + 3
E = ( x - 1)2 + ( y + 2)2 + 3
Do : ( x - 1)2 lớn hơn hoặc bằng 0 với mọi x
( y + 2)2 lớn hơn hoặc bằng 0 với mọi x
Suy ra : ( x - 1)2 + 3 lớn hơn hoặc bằng 3 với mọi x
( y + 2)2 + 3 lớn hơn hoặc bằng 3 với mọi x
Vậy , Emin = 3 khi và chỉ khi x - 1 =0 -> x = 1
y + 2 =0 -> y = -2
b) F = x2 - 4x + y2 - 8y + 6
F = x2 - 4x + y2 - 8y + 4 + 16 - 14
F = ( x2 - 2.2x + 22) + ( y2 - 2.4y + 42) - 14
F = ( x - 2)2 + ( y - 4)2 - 14
Do : ( x - 2)2 lớn hơn hoặc bằng 0 với mọi x
( y - 4)2 lớn hơn hoặc bằng 0 với mọi x
Suy ra : ( x - 2)2 - 14 lớn hơn hoặc bằng -14 với mọi x
( y - 4)2 -14 lớn hơn hoặc bằng -14 với mọi x
Vậy , Fmin = -14 khi và chỉ khi x - 2 =0 -> x = 2
y - 4 = 0 -> y = 4
Bài 2 . a) 3x2 - 3y2 - 2( x - y)2
= 3( x - y)(x + y) - 2( x - y)( x - y)
= (x - y)( 3x + 3y - 2x + 2y)
b) x3 - 4x2 - 9x + 36
= x2(x - 4) - 9( x - 4)
= ( x - 4)( x2 - 32)
= ( x - 4)( x - 3)( x + 3)
c) 3x2 - 6xy + 3y2 - 12z2
= 3( x2 - 2xy + y2 - 4z2)
= 3[( x - y)2 - ( 2z)2]
= 3( x - y - 2z)( x - y + 2z)
d) 5x2 - 10xy + 5y2 - 20x2
= 5( x2 - 2xy + y2 - 4x2)
= 5[ ( x - y)2 - ( 2x)2 ]
= 5( x - y - 2x)( x - y + 2x)
2.
a) 3x2 - 3y2 - 2(x - y)2
= 3(x2 - y2) - 2(x - y)2
= 3(x - y)(x + y) - 2(x - y)2
= [3(x + y) - 2(x - y)](x - y)
= (3x + 3y - 2x + 2y)(x - y)
= (x + 5y)(x - y)
b) x3 - 4x2 - 9x + 36
= x2(x - 4) - 9(x - 4)
= (x2 - 9)(x - 4)
c) 3x2 - 6xy + 3y2 - 12z2
= 3(x2 - 2xy + y2) - 3.4z2
= 3(x - y)2 - 3(2z)2
= 3[(x - y)2 - (2z2)]
= 3(x - y + 2z)(x - y - 2z)
d) 5x2 - 10xy + 5y2 - 20x2
= 5(x2 - 2xy + y2) - 5.4x2
= 5(x - y)2 - 5(2x)2
= 5(x - y + 2x)(x - y - 2x)
Câu 1:
\(\text{ a) }E=x^2-2x+y^2+4y+8\\ E=x^2-2x+y^2+4y+1+4+3\\ E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\\ E=\left(x-1\right)^2+\left(y+2\right)^2+3\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \left(y+2\right)^2\ge0\forall y\\ \Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x;y\\ \Rightarrow E=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x;y\\ \text{ Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \text{ Vậy }E_{\left(Min\right)}=3\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(\text{b) }F=x^2-4x+y^2-8y+6\\ F=x^2-4x+y^2-8y+4+16-14\\ F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \left(y-4\right)^2\ge0\forall y\\ \left(x-2\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ F=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x;y\\ \text{Dấu “=” xảy ra khi : }\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\\ \text{Vậy }F_{\left(Min\right)}=-14\text{ }khi\text{ }\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
Câu 2:
\(\text{a) }3x^2-3y^2-2\left(x-y\right)^2\\ =\left(3x^2-3y^2\right)-2\left(x-y\right)^2\\ =3\left(x^2-y^2\right)-2\left(x-y\right)^2\\ =3\left(x+y\right)\left(x-y\right)-2\left(x-y\right)^2\\ =\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)
\(\text{b) }x^2-4x^2-9x+36\\ =x^3-7x^2+3x^2+12x-21x+36\\ =\left(x^3-7x^2+12x\right)+\left(3x^2-21x+36\right)\\ =x\left(x^2-7x+12\right)+3\left(x^2-7x+12\right)\\ =\left(x+3\right)\left(x^2-7x+12\right)\\ =\left(x+3\right)\left(x^2-4x-3x+12\right)\\ =\left(x+3\right)\left[\left(x^2-4x\right)-\left(3x-12\right)\right]\\ =\left(x+3\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]\\ =\left(x+3\right)\left(x-3\right)\left(x-4\right)\)
\(\text{c) }3x^2-6xy+3y^2-12z^2\\ =3\left(x^2-2xy+y^2-4z^2\right)\\ =3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\\ =3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\\ =3\left(x-y+2z\right)\left(x-y-2z\right)\)
\(\text{d) }5x^2-10xy+5y^2-20x^2\\ =-15x^2+10xy+5y^2\\ =-5\left(3x^2-2xy-y^2\right)\\ =-5\left(3x^2-3xy+xy-y^2\right)\\ =-5\left[\left(3x^2-3xy\right)+\left(xy-y^2\right)\right]\\ =-5\left[3x\left(x-y\right)+y\left(x-y\right)\right]\\ =-5\left(3x+y\right)\left(x-y\right)\)
1)
a) \(E=x^2-2x+y^2+4y+8\)
\(\Leftrightarrow E=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)
\(\Leftrightarrow E=\left(x-1\right)^2+\left(y+2\right)^2+3\)
Vậy GTNN của E=3 khi \(\left\{{}\begin{matrix}x-1=0\Leftrightarrow x=1\\y+2=0\Leftrightarrow y=-2\end{matrix}\right.\)
b) \(F=x^2-4x+y^2-8y+6\)
\(\Leftrightarrow F=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)
\(\Leftrightarrow F=\left(x-2\right)^2+\left(y-4\right)^2-14\)
Vậy GTNN của \(F=-14\) khi \(\left\{{}\begin{matrix}x-2=0\Leftrightarrow x=2\\y-4=0\Leftrightarrow y=4\end{matrix}\right.\)
2)
a)\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left[3x+3y-2x+2y\right]\)
\(=\left(x-y\right)\left(x+5y\right)\)
b) \(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
c) \(3x^2-6xy+3y^2-12z^2\)
\(=3\left(x^2-2xy+y^2-4z^2\right)\)
\(=3\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
d) \(5x^2-10xy+5y^2-20x^2\)
\(=5\left(x^2-2xy+y^2-4x^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-4x^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=5\left(x-y-2x\right)\left(x-y+2x\right)\)
\(=5\left(-x-y\right)\left(3x-y\right)\)
