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ta có \(A=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)
\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)
vì \(-\left(2x+1\right)^2< =0;-\left(3y-1\right)^2< =0\)
=> \(A< =5\)
dấu = xảy ra <=> \(\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{3}\end{cases}}\)
b) ta có \(B=-\left(x^2-6x-5\right)=-\left[\left(x^2-6x+9\right)-14\right]\)
\(=-\left(x-3\right)^2+14\)
mà \(-\left(x-3\right)^2< =0\) => b<=14
dấu = xảy ra <=> \(x=3\)
a/ \(A=x^2+y^2-2x+6y+12\)
\(=\left(x^2-2x+1\right)+\left(y^2+6y+9\right)+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+3\right)^2\ge0\)
\(\Leftrightarrow A\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
Vậy....
b/ \(B=-4x^2-9y^2-4x+6y+3\)
\(=-\left(4x^2+4x+1\right)-\left(9y^2+6y+1\right)+1\)
\(=-\left(2x+1\right)^2-\left(3y+1\right)^2+1\)
Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(2x+1\right)^2\ge0\\\left(3y+1\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0\\-\left(3y+1\right)^2\le0\end{matrix}\right.\)
\(\Leftrightarrow-\left(2x+1\right)^2-\left(3y+1\right)^2\le0\)
\(\Leftrightarrow B\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-\frac{1}{3}\end{matrix}\right.\)
Ta có : P = x2 - 2x + 5 = x2 - 2x + 1 + 4 = (x - 1)2 + 4
Vì \(\left(x-1\right)^2\ge0\forall x\)
Suy ra : \(P=\left(x-1\right)^2+4\ge4\forall x\)
Nên : Pmin = 4 khi x = 1
b) Ta có Q = 2x2 - 6x = 2(x2 - 3x) = 2(x2 - 3x + \(\frac{9}{4}-\frac{9}{4}\) ) = \(2\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)
Vì \(2\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
SUy ra ; \(Q=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Vậy \(Q_{min}=-\frac{9}{2}\) khi \(x=\frac{3}{2}\)
\(a)\)
\(A=2x^2+x\)
\(\Leftrightarrow A=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
\(MinA=\frac{-1}{8}\)khi \(x=\frac{-1}{4}\)
\(b)\)
\(B=x^2+2x+y^2-4y+6\)
\(\Leftrightarrow B=x^2+2x+1+y^2-4y+4+1\)
\(\Leftrightarrow B=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu '' = '' xảy ra khi: \(x=-1;y=2\)
\(c)\)
\(C=4x^2+4x+9y^2-6y-5\)
\(\Leftrightarrow C=4x^2+4x+1+9y^2-6y+1-7\)
\(\Leftrightarrow C=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
Dấu '' = '' xáy ra khi: \(x=\frac{-1}{2};y=\frac{1}{3}\)
C =- (4x2+4x+1) - (9y2 -6y +1) +3 = - (2x+1)2 - ( 3y -1)2 + 3 </ 3
C max = 3 khi x =-1/2 và y =1/3
D - dể suy nghĩ đã nhé
b: Ta có: \(B=x^2+4x+9y^2-6y-1\)
\(=x^2+4x+4+9y^2-6y+1-6\)
\(=\left(x+2\right)^2+\left(3y-1\right)^2-6\ge-6\forall x,y\)
Dấu '=' xảy ra khi x=-2 và \(y=\dfrac{1}{3}\)
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3