Tìm GTNN của:
a) A=x2+2y2-6x+8y+25
b)B=x2+3y2+2xy-2x-10y+19
c)C=4x2-2xy+y2-14x-4y+50
d)D=2x2+2y2+2xy-8x-8y+20
Làm được câu nào thì làm, thưởng từ 6sp trở lên~
tìm gtnn (gtln) của:
a) A= 4x2-4x+10 b) B= 2x2-3x-1
c) C= 4x2+2y2+4xy+4x+6y+1 d) D= (3x-1)2-4(3x-1)x+4x2
e) G= 9x2+2y2+6xy+4y+5 f) H= 2x2+3y2-2xy+4y+2x+5
g) K= xy+yz+zx; biết x+y+z= 3
nhờ mn giúp mik vs nha
\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)
\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)
\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)
Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)
tìm GTNN của các bt
a, A=2x2+y2-2xy-2x+3
b,B=x2-2xy+2y2+2x-10y+17
c,C=x2-xy+y2-2y-2x
d,D=x2+xy+y2-3y-3x
e,E=2x2+2xy +5y2-8x-22y
A= 2x^2 + y^2 - 2xy -2x+3
A= x^2-2xy + y^2 + x^2 - 2x+ 1 +2
A= (x-y)^2 + (x-1)^2 + 2
(x-y)^2> hoặc = 0 với mọi giá trị của x
(x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 > hoặc =0 với mọi giá trị của x
=> (x-y)^2 + (x-1)^2 + 2 > hoặc =2
=> A lớn hơn hoặc bằng 2
=> GTNN của A=2 tại x=y=1
Giúp me zới!!!
Bài 1: Tìm giá trị nhỏ nhất:
a)A=x2-2xy+5y2+4y+51
b)B=121/-4xy2-12x+2
c)C=9/-2x2+4x-7
d)10x2+4y2-4xy+8x-4y+20
e)E=9x2+2y2+6xy-6x-8y+10
a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)
\(=x^2-2xy+y^2+4y^2+4y+1+50\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)
a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)
\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)
c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)
\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)
d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)
\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)
\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)
A=-x2+3xy+2y2;B=4x2-5xy+3y2;C=3x2+2xy+y2.Tính A-B-C
\(A-B-C=\left(-x^2+3xy+2y^2\right)-\left(4x^2-5xy+3y^2\right)-\left(3x^2+2xy+y^2\right)\)
\(=-x^2+3xy+2y^2-4x^2+5xy-3y^2-3x^2-2xy-y^2\)
\(=-8x^2+6xy-2y^2\)
Tìm GTLN của BT sau
-x2+3x
-5x2-2xy-2y2+14x+10y-1
-8x2-3y2-26x+6y+100
\(-5x^2-2xy-2y^2+14x+10y-1\\ =-\left(x^2+2xy+y^2\right)-\left(4x^2-2\cdot2\cdot\dfrac{7}{2}x+\dfrac{49}{4}\right)-\left(y^2-10y+25\right)+\dfrac{55}{4}\\ =-\left(x+y\right)^2-\left(2x-\dfrac{7}{2}\right)^2-\left(y-5\right)^2+\dfrac{55}{4}\le\dfrac{55}{4}\\ Max\Leftrightarrow\left\{{}\begin{matrix}x=-y\\2x=\dfrac{7}{2}\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=\dfrac{7}{4}\\y=5\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
a: Ta có: \(-x^2+3x\)
\(=-\left(x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
Tìm GTNN:
a) B= x2 + 2y2 - 2xy - 4y + 5
b) C= 2x2 - 2xy + 5y2 +5
Cho các đa thức A = 4 x 2 - 5 x y + 3 y 2 ; B = 3 x 2 + 2 x y + y 2 ; C = - x 2 + 3 x y + 2 y 2 . Tính A + B + C
A. 7 x 2 + 6 y 2
B. 5 x 2 + 5 y 2
C. 6 x 2 + 6 y 2
D. 6 x 2 - 6 y 2
Cho các đa thức A = 4 x 2 - 5 x y + 3 y 2 ; B = 3 x 2 + 2 x y + y 2 ; C = - x 2 + 3 x y + 2 y 2 . Tính A - B - C
A. - 10 x 2 + 2 x y
B. - 2 x 2 - 10 x y
C. 2 x 2 + 10 x y
D. 2 x 2 - 10 x y
Cho các đa thức A = 4 x 2 - 5 x y + 3 y 2 ; B = 3 x 2 + 2 x y + y 2 ; C = - x 2 + 3 x y + 2 y 2
Tính C - A - B
A. 8 x 2 + 6 x y + 2 y 2
B. - 8 x 2 + 6 x y - 2 y 2
C. 8 x 2 - 6 x y - 2 y 2
D. 8 x 2 - 6 x y + 2 y 2
Ta có
C − A − B = − x 2 + 3 x y + 2 y 2 − 4 x 2 − 5 x y + 3 y 2 − 3 x 2 + 2 x y + y 2 = − x 2 + 3 x y + 2 y 2 − 4 x 2 + 5 x y − 3 y 2 − 3 x 2 − 2 x y − y 2 = − x 2 − 4 x 2 − 3 x 2 + ( 3 x y + 5 x y − 2 x y ) + 2 y 2 − 3 y 2 − y 2 = − 8 x 2 + 6 x y − 2 y 2
Chọn đáp án B