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a) \(x^2+xy+y^2+1\)
\(=x^2+xy+\dfrac{y^2}{4}-\dfrac{y^2}{4}+y^2+1\)
\(=\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1\)
mà \(\left\{{}\begin{matrix}\left(x+\dfrac{y}{2}\right)^2\ge0,\forall x;y\\\dfrac{3y^2}{4}\ge0,\forall x;y\end{matrix}\right.\)
\(\Rightarrow\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}+1>0,\forall x;y\)
\(\Rightarrow dpcm\)
b) \(...=x^2-2x+1+4\left(y^2+2y+1\right)+z^2-6z+9+1\)
\(=\left(x-1\right)^2+4\left(y^{ }+1\right)^2+\left(z-3\right)^2+1>0,\forall x.y\)
\(\Rightarrow dpcm\)
x2+y2+z2+2x-4y+6z=-14
=>x2+y2+z2+2x-4y+6z+14=0
=>(x2+2x+1)+(y2-4y+4)+(z2+6z+9)=0
=> (x+1)2+(y-2)2+(z+3)2=0
ta có:
(x+1)2≥0
(y-2)2≥0
(z+3)2≥0
=>(x+1)2+(y-2)2+(z+3)2≥0
dấu "=" xảy ra khi và chỉ khi ; x+1=y-2=z+3=0
=>\(\hept{\begin{cases}x=-1\\y=2\\z=-3\end{cases}}\)
=> x+y+z=-1+2+(-3)=-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
$A=x^2+y^2-6x+4y+20=(x^2-6x+9)+(y^2+4y+4)+7$
$=(x-3)^2+(y+2)^2+7\geq 0+0+7=7$
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 $x+y+1=x-1=0$
$\Leftrightarrow x=1; y=-2$
\(Q=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2x^2+2y^2}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2+2y^2+4xy}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)
điều kiện ban đầu <=> (x-1)2+(y-2)2+(z-3)2 \(\le1\)
áp dụng bdt sau (ax+ by+ cz)2\(\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)(bunhiacopxky với 3 số)
[ x-1 + 2(y-2) + 2(z-3)]2 \(\le\left(1^2+2^2+2^2\right)\left[\left(x-1\right)^2+\left(y-2\right)^2+\left(z-2\right)^2\right]\le9.1=9\)
=>\(-3\le\) x-1 +2(y-2) +2(z-3) \(\le3\) <=> 8\(\le x+2y+2z\le14\)
\(x^2+y^2+z^2+2x-4y-6z+14\)
\(=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2-6z+9\right)\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\)
Vì \(\left(x+1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\); \(\left(z-3\right)^2\ge0\forall z\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z-3\right)^2\ge0\forall x,y,z\)
hay \(x^2+y^2+z^2+2x-4y-6z+14\ge0\)\(\forall x,y,z\)
Ta có : \(A=x^2+y^2+z^2-2x-4y+6z=-14\)
\(\Leftrightarrow x^2+y^2+z^2+2x-4y+6z+14=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6z+9\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\left(1\right)\)
Do \(\left(x+1\right)^2\ge0\forall x;\left(y-2\right)^2\ge0\forall y;\left(z+3\right)^2\ge0\forall z\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2\ge0\forall x;y;z\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\\\left(z+3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\y-2=0\\z+3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\\z=-3\end{matrix}\right.\)
\(\Rightarrow x+y+z=-1+2-3=-2\)
Vậy \(x+y+z=-2\)