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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
\(\left(x^2+4z\right)^2=17\left(x^4+z^2\right)\)
\(x^4+8x^2z+16z^2=17x^4+17z^2\)
\(t^4-2t^2z+z^2=\left(t^2-z\right)^2=0\)
Nghiệm duy nhất: \(t^2=z\Rightarrow t^2=y^2+7\Rightarrow\hept{\begin{cases}t=4\Rightarrow x=2\\y=3\end{cases}}\)KL (x,y)=(2,3)
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
a) \({y^2} + y + \dfrac{1}{4} = {y^2} - 2.y.\dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} = {\left( {y - \dfrac{1}{2}} \right)^2}\)
b) \({y^2} + 49 - 14y = {y^2} - 14y + 49 = {y^2} - 2.y.7 + {7^2} = {\left( {y - 7} \right)^2}\)
\(\left(x^2+4y^2+28\right)^2=17\left[x^4+\left(y^2+7\right)^2\right]\)
y^2 +7 =z
\(\Leftrightarrow x^4+8xz+16z^2=17x^4+17z^2\)
\(\Leftrightarrow16x^4+z^2-8xz=0\)\(\Leftrightarrow\left(4x^2-z\right)^2=0\)
\(\Leftrightarrow4x^2=z\Leftrightarrow4x^2-y^2=7\)
\(\left\{{}\begin{matrix}4x^2=16\\y^2=9\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(\pm2;\pm3\right)\)
\(\dfrac{y^2-14y-1}{y^2-4y+4}-y^2-6y\)
\(\Leftrightarrow\dfrac{y^2-14y-1}{y^2-4y+4}-\dfrac{\left(y^2+6y\right)\left(y^2-4y+4\right)}{y^2-4y+4}\)
\(\Rightarrow y^2-14y-1-\left(y^2+6y\right)\left(y^2-4y+4\right)\)
\(\Rightarrow\)
y2-14y-1-(y4-4y3+4y2+6y3-24y2+24y)
\(\Rightarrow\)y2-14y-1-y4+4y3-4y2-6y3+24y2-24y
\(\Rightarrow\)-y4-2y3+21y2-38y-1