X^2+2y^2+z^2-2(xy+2y+2z+8)=0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
0
NV
0
NV
0
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 5 2020
\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}\)
\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}\)
\(VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)\)
\(VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
G
8 tháng 7 2019
\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)
Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:
\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)
\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)
15 tháng 5 2017
a) x6+x2y5+xy6+x2y5-xy6
= x6+(x2y5+x2y5)+(xy6-xy6)
= x6+2x2y5
b) \(\dfrac{1}{2}\)x2y3-x2y3+3x2y2z2-z4-3x2y2z2
= (\(\dfrac{1}{2}\)x2y3-x2y3)+(3x2y2z2-3x2y2z2)-z4
= -\(\dfrac{1}{2}\)x2y3-z4
\(x^2+2y^2+z^2-2\left(xy+2y+2z+8\right)=0\)
\(pt\Leftrightarrow x^2+2y^2+z^2-2xy+4y+4z+16=0\)
\(\Leftrightarrow x^2-2xy+y^2+y^2+4y+4+z^2+4z+4+8=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8=0\)
Dễ thấy: \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y+2\right)^2\ge0\\\left(z+2\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2+\left(z+2\right)^2+8>8\)
Vô nghiệm