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\(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Rightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Rightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có : \(\left(x-1\right)^2\ge0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(2y-3\right)^2\ge0\Rightarrow2y-3=0\Rightarrow2y=3\Rightarrow y=\frac{3}{2}\)
\(\left(z+2\right)^2\ge0\Rightarrow z+2=0\Rightarrow z=-2\)
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\(=\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow x=1;y=\frac{3}{2};z=-2\)
Ta có:
x2+4y2+z2-2x-12y-4z-14=0
x2-2x+1+z2-4z+4+4y2-12y+9=0
(x-1)2+(z-2)2+(2y-3)2=0
Tổng 3 số không âm bằng 0
<=> x-1=0 và z-2=0 và 2y-3=0
<=> x=1 và z=2 và y=3/2
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a. \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
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Bài 1:
x3+y3=152=> (x+y)(x2-xy+y2)=152
Mà x2-xy+y2=19
=> 19(x+y)=152=> x+y=8
Ta cũng có x-y=2
=> x=5;y=3
Bài 2:
x2+4y2+z2=2x+12y-4z-14
=> x2+4y2+z2-2x-12y+4z+14=0
=> (x2-2x+1)+(4y2-12y+9)+(z2+4z+4)=0
=> (x+1)2+(2y-3)2+(z+2)2=0
=> (x+1)2=(2y-3)2=(z+2)2=0
=> x=-1;y=3/2;z=-2
Bài 3\(\left(\frac{1}{x^2+x}-\frac{1}{x+1}\right):\frac{1-2x+x^2}{2014x}=\left(\frac{1}{x\left(x+1\right)}-\frac{1}{x+1}\right):\frac{\left(1-x\right)^2}{2014x}=\frac{1-x}{x\left(x+1\right)}.\frac{2014x}{\left(1-x\right)^2}=\frac{2014}{\left(x+1\right)\left(1-x\right)}=\frac{2014}{1-x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+9y^2+4z^2-2x+12y-4z+20=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(9y^2+12y+4\right)+\left(4z^2-4z+1\right)+14=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14=0\)(1)
Ta thấy\(\left(x-1\right)^2+\left(3y+2\right)^2+\left(2z-1\right)^2+14\ge14>0\forall x;y;z\)
Nên dấu (1) không thể xảy ra , Hay \(x;y;z\) ko tồn tại (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
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
Lời giải:
$x^2+4y^2+z^2=2x+12y-4z-14$
$\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0$
$\Leftrightarrow (x^2-2x+1)+(4y^2-12y+9)+(z^2+4z+4)=0$
$\Leftrightarrow (x-1)^2+(2y-3)^2+(z+2)^2=0$
Vì $(x-1)^2\geq 0; (2y-3)^2\geq 0; (z+2)^2\geq 0$ với mọi $x,y,z\in\mathbb{R}$
Do đó để tổng của chúng bằng $0$ thì:
$(x-1)^2=(2y-3)^2=(z+3)^2=0$
$\Rightarrow x=1; y=\frac{3}{2}; z=-3$