Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = -x2 + 2xy - 4y2 + 2x + 10y - 8
=> -A = x2 - 2xy + 4y2 - 2x - 10y + 8
= ( x2 - 2xy + y2 - 2x + 2y + 1 ) + ( 3y2 - 12y + 12 ) - 5
= [ ( x2 - 2xy + y2 ) - ( 2x - 2y ) + 1 ] + 3( y2 - 4y + 4 ) - 5
= [ ( x - y )2 - 2( x - y ) + 1 ] + 3( y - 2 )2 - 5
= ( x - y - 1 )2 + 3( y - 2 )2 - 5 ≥ -5 ∀ x, y
Dấu "=" xảy ra <=> x = 3 ; y = 2
=> -A ≥ -5
=> A ≤ 5
=> MaxA = 5 <=> x = 3 ; y = 2
B = 2x2 + 9y2 - 6xy - 6x - 12y + 2004
= ( x2 - 6xy + 9y2 + 4x - 12y + 4 ) + ( x2 - 10x + 25 ) + 1975
= [ ( x2 - 6xy + 9y2 ) + ( 4x - 12y ) + 4 ] + ( x - 5 )2 + 1975
= [ ( x - 3y )2 + 2( x - 3y ).2 + 22 ] + ( x - 5 )2 + 1975
= ( x - 3y + 2 )2 + ( x - 5 )2 + 1975 ≥ 1975 ∀ x, y
Dấu "=" xảy ra <=> x = 5 ; y = 7/3
=> MinB = 1975 <=> x = 5 ; y = 7/3
Ta có: A = -x2 + 2xy - 4y2 + 2x + 10y - 8
A = -[x2 - 2xy + 4y2 - 2x - 10y + 8]
A = -[(x2 - 2xy + y2) - 2(x + y) + 1 + 3y2 - 12y + 12 - 5]
A = -[(x - y)2 - 2(x + y) + 1 + 3(y - 2)2]+ 5
A = -[(x - y - 1)2 + 3(y - 2)2] + 5 \(\le\) 5 với mọi x
Dấu "=" xảy ra <=> x - y - 1 = 0 và y + 2 = 0
=>x = -1 và y = -2
Vậy MaxA = 5 khi x = -1 và y = -2
B = 2x2 + 9y2 - 6xy - 6x - 12y + 2004
B = (x2 - 6xy + 9y2) + 4(x - 3y) + 4 + x2 - 10x + 25 + 1975
B = (x - 3y + 2)2 + (x - 5)2 + 1975 \(\ge\)1975
đoạn cuối tt trên
\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
A= 4x2 - 3x + 1
= (2x) 2 - 2.2x.4/3 + (4/3) 2 - (4/3) 2 + 1
= (2x - 4/3) 2 - 7/9
Nhận xét: (2x - 4/3) 2 \(\ge\)0 với mọi x
=> (2x - 4/3) 2 - 7/9 \(\le\) 7/9
=> Min A là 9
Dấu "=" xảy ra <=> 2x - 4/3 = 0 <=> 2x = 4/3 <=> x = 2/3
Vậy..
a) \(25.\left(x-1\right)^2-16\left(x+y\right)^2\)
= \(\left(5x-5\right)^2-\left(4x+y\right)^2\)
= \(\left(5x-5-4x-y\right)\left(5x-5+4x+y\right)\)
= \(\left(x-y-5\right)\left(9x+y-5\right)\)
b) \(x^3+3x^2+3x+1-27z^3\)
= \(\left(x+1\right)^3-27z^3\)
= \(\left(x+1-3z\right)\left(x^2+x.3z+9z^2\right)\)
c) \(x^2-2xy+y^2-xz+yz\)
= \(\left(x-y\right)^2-z\left(x-y\right)\)
= \(\left(x-y\right)\left(x-y-z\right)\)
d) \(a^3x-ab+b-x\)
= \(x\left(a^3-1\right)-b\left(a-1\right)\)
= \(x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
= \(\left(a-1\right)\left(a^2x+ax+x-b\right)\)
f) \(x^2+2x-4y^2-4y\)
= \(x^2+2x+1-\left(4y^2+4y+1\right)\)
= \(\left(x+1\right)^2-\left(2y+1\right)^2\)
= \(\left(x+1-2y-1\right)\left(x+1+2y+1\right)\)
= \(\left(x-2y\right)\left(x+2y+2\right)\)
g) \(xy-4+2x-2y\)
= \(y\left(x-2\right)-2\left(x-2\right)\)
= \(\left(x-2\right)\left(y-2\right)\)
a: \(=\left(5x-5\right)^2-\left(4x-4y\right)^2\)
\(=\left(5x-5-4x+4y\right)\cdot\left(5x-5+4x-4y\right)\)
\(=\left(x+4y-5\right)\left(9x-4y-5\right)\)
b: \(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
c: \(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
d: \(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\cdot\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2x+ax+1-b\right)\)
a) \(A=2x^2+9y^2-6xy-6x-12y+2014\)
\(=\left(2x^2-6xy-6x\right)+\left(9y^2-12y\right)+2014\)
\(=2\left[x^2-2.x.\frac{3\left(y+1\right)}{2}+\frac{9\left(y+1\right)^2}{4}\right]+\left[9y^2-12y-\frac{9}{2}.\left(y+1\right)^2\right]+2014\)
\(=2\left[x-\frac{3\left(y+1\right)}{2}\right]^2+\frac{1}{2}\left(3y-7\right)^2+1985\ge1985\)
Dấu "=" xảy ra khi và chỉ khi y = \(\frac{7}{3}\Rightarrow x=5\)
Vậy Min A = 1985 tại \(\left(x;y\right)=\left(5;\frac{7}{3}\right)\)
b) \(B=-x^2+2xy-4y^2+2x+10y-8\)
\(=-\left(x^2-2xy-2x\right)-\left(4y^2-10y\right)-8\)
\(=-\left[x^2-2x\left(y+1\right)+\left(y+1\right)^2\right]-\left[4y^2-10y-\left(y+1\right)^2\right]-8\)
\(=-\left(x-y-1\right)^2-\left(y-2\right)^2+5\le5\)
Dấu đẳng thức xảy ra khi và chỉ khi y = 2 => x = 3
Vậy B đạt giá trị lớn nhất bằng 5 tại (x;y) = (3;2)