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\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
b: \(8x^2-48x+6xy-36y\)
\(=8x\left(x-6\right)+6y\left(x-6\right)\)
\(=2\left(x-6\right)\left(4x+3y\right)\)
d: \(a^2-2ab+b^2-4\)
\(=\left(a-b\right)^2-4\)
\(=\left(a-b-2\right)\left(a-b+2\right)\)
$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$
$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
Vậy $A_{\min}=7$. Giá trị này đạt tại $(x-3)^2=(y+2)^2=0$
$\Leftrightarrow x=3; y=-2$
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$B=9x^2+y^2+2z^2-18x+4z-6y+30$
$=(9x^2-18x+9)+(y^2-6y+9)+(2z^2+4z+2)+10$
$=9(x^2-2x+1)+(y^2-6y+9)+2(z^2+2z+1)+10$
$=9(x-1)^2+(y-3)^2+2(z+1)^2+10\geq 10$
Vậy $B_{\min}=10$. Giá trị này đạt tại $(x-1)^2=(y-3)^2=(z+1)^2$
$\Leftrightarrow x=1; y=3; z=-1$
$C=x^2+y^2+z^2-xy-yz-xz+3$
$2C=2x^2+2y^2+2z^2-2xy-2yz-2xz+6$
$=(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)+6$
$=(x-y)^2+(y-z)^2+(z-x)^2+6\geq 6$
$\Rightarrow C\geq 3$
Vậy $C_{\min}=3$. Giá trị này đạt tại $x-y=y-z=z-x=0$
$\Leftrihgtarrow x=y=z$
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$D=5x^2+2y^2+4xy-2x+4y+2021$
$=2(y^2+2xy+x^2)+3x^2-2x+4y+2021$
$=2(x+y)^2+4(x+y)+3x^2-6x+2021$
$=2(x+y)^2+4(x+y)+2+3(x^2-2x+1)+2016$
$=2[(x+y)^2+2(x+y)+1]+3(x^2-2x+1)+2016$
$=2(x+y+1)^2+3(x-1)^2+2016\geq 2016$
Vậy $D_{\min}=2016$ khi $x+y+1=x-1=0$
$\Leftrightarrow x=1; y=-2$
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
\(A=-4x^2-5y^2+8xy+10y+12\)
\(-A=4x^2+5y^2-8xy-10y-12\)
\(-A=\left(4x^2-8xy+y^2\right)+\left(4y^2-10y+\frac{25}{4}\right)-\frac{73}{4}\)
\(-A=\left(2x-y\right)^2+\left(2y-\frac{5}{2}\right)^2-\frac{73}{4}\)
Mà : \(\left(2x-y\right)^2\ge0\forall x;y\)
\(\left(2y-\frac{5}{2}\right)^2\ge0\forall y\)
\(\Rightarrow-A\ge-\frac{73}{4}\)
\(\Leftrightarrow A\le\frac{73}{4}\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}2x-y=0\\2y-\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{8}\\y=\frac{5}{4}\end{cases}}\)
Vậy \(A_{Max}=\frac{73}{4}\Leftrightarrow\left(x;y\right)=\left(\frac{5}{8};\frac{5}{4}\right)\)