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D
1
ND
27 tháng 6 2018
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. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
20 tháng 1 2020
Chia nhỏ ra bạn ơi!
\(a) x² +3y²+2z²-2x+12y+4z+15=0 \)
\(⇔x²-2x+1+3y²+12y+12+2z²+4z+2=0 \)
\(⇔(x²-2x+1) + 3(y²+4y+4) +2(z²+2z+1)=0 \)
\(⇔(x-1)² +3(y+2)²+2(z+1)²=0 \)
\(⇔ x-1=0 \) và \(y+2=0\) và \(z+1=0\)
Vậy: \(x=1;y=-2;z=-1\)
NM
0
4 tháng 8 2017
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
ta có : \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\) \(\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\\\left(z+1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
vậy \(x=1;y=-2;z=-1\)
4 tháng 8 2017
\(x^2+3y^2+2z^2-2z+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(3y^2+12y+12\right)+\left(2z^2-4z+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+4\right)^2+2\left(z-2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-4\\z=2\end{matrix}\right.\)
5 tháng 8 2017
Bài 1:
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow x^2-2x+1+3y^2+12y+12+2z^2+4z+2=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
Dễ thấy: \(\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\3\left(y+2\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
Bài 2:
a)\(A=x^2-4xy+5y^2+10x-22y+28\)
\(=x^2-4xy+10x+4y^2-20y+25+y^2-2y+1+2\)
\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)
b)\(B=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+15\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+15\)
Đặt \(t=x^2-5x+4\) thì ta có:
\(t\left(t+2\right)+15=t^2+2t+1+14\)
\(=\left(t+1\right)^2+14\ge14\)
Xảy ra khi \(t=-1 \)\(\Rightarrow x^2-5x+4=-1\Rightarrow x=\dfrac{5\pm\sqrt{5}}{2}\)
QN
1
LT
30 tháng 12 2018
Ta có 2x=3y =>\(\dfrac{x}{3}=\dfrac{y}{2}\)=>\(\dfrac{x}{12}=\dfrac{y}{8}\)
5y=4z=>\(\dfrac{y}{4}=\dfrac{z}{5}\)=>\(\dfrac{y}{8}=\dfrac{z}{10}\)
=>\(\dfrac{x}{12}=\dfrac{y}{8}=\dfrac{z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được
\(\dfrac{x^2}{144}=\dfrac{y^2}{64}=\dfrac{z^2}{100}=\)\(\dfrac{x^2}{144}=\dfrac{3y^2}{192}=\dfrac{2z^2}{200}\)
=\(\dfrac{342}{152}=\dfrac{9}{4}\)
Còn lại tự làm nha
HN
28 tháng 4 2017
Ta đặt: \(\left\{{}\begin{matrix}\dfrac{1}{x^2}=a\\\dfrac{1}{y^2}=b\\\dfrac{1}{z^2}=c\end{matrix}\right.\)\(\Rightarrow\sqrt{abc}=abc=1\)
Ta có: \(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\)
\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\dfrac{1}{\sqrt{a}}+1}+\dfrac{1}{\dfrac{1}{\sqrt{ab}}+\sqrt{ca}+1}\)
\(=\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{\sqrt{a}}{\sqrt{ba}+1+\sqrt{a}}+\dfrac{1}{1+\sqrt{ab}+\sqrt{a}}=1\)
Quay lại bài toán, sau khi đặt bài toán trở thành:
\(P=\dfrac{1}{2b+a+3}+\dfrac{1}{2c+b+3}+\dfrac{1}{2a+c+3}\)
\(=\dfrac{1}{\left(a+b\right)+\left(b+1\right)+2}+\dfrac{1}{\left(b+c\right)+\left(c+1\right)+2}+\dfrac{1}{\left(c+a\right)+\left(a+1\right)+2}\)
\(\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}+\sqrt{ab}+1}+\dfrac{1}{\sqrt{b}+\sqrt{bc}+1}+\dfrac{1}{\sqrt{c}+\sqrt{ca}+1}\right)=\dfrac{1}{2}\)
HN
28 tháng 4 2017
Cái đó t cố tình bỏ đấy. B phải tự làm chứ chẳng lẽ t làm hết??
6 tháng 9 2021
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
6 tháng 9 2021
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)
\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)
\(\Rightarrow x=1;y=-2;z=-1\)
<=>(x2-2x+1)+(3y2+12y+12)+(2z2+4z+2)=0
<=>(x-1)2+3(y+2)2+2(z+1)2=0
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\3\left(y+2\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)