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 \(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-\frac{1}{3}\end{matrix}\right.\)

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI

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 : P_min = 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

ai ủng hộ vài li-ke tròn 210 lun , please