Cho \(x^2y-xy^2+x^2z-xz^2+y^2z+yz^2=2xyz\). CMR: trong 3 số \(x,y,z\) có ít nhất hai số bằng nhau hoặc đối nhau.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
(x2 y - y2 x) + (x2 z - xyz) + (z2 y - z2 x) + (y2 z - xyz) = (x-y)(xy+zx-z2 -yz)=(x-y)(x-z)(y+z)=0
Giải giùm rồi đấy bạn
\(\Leftrightarrow\left(x^2y-2xyz+z^2y\right)+\left(x^2z-y^2x-z^2x+y^2z\right)=0\)
\(\Leftrightarrow y\left(x-z\right)^2+xz\left(x-z\right)-y^2\left(x-z\right)=0\)
\(\Leftrightarrow\left(x-z\right)\left(xy-yz+zx-y^2\right)=0\)
\(\Leftrightarrow\left(x-z\right)\left(x\left(y+z\right)-y\left(y+z\right)\right)=0\)
\(\Leftrightarrow\left(x-z\right)\left(x-y\right)\left(y+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\) hay có 2 số bằng hoặc đối nhau
Bài này bạn phải chuyển 2xyz sang vế kia rồi nhóm hợp lí mới ra được
(x2y+z2y-2xyz)-(y2x-y2z)+(x2z-z2x)=0
y(x2+z2-2xz)-y2(x-z)+xz(x-z)=0
y(x-z)(x-z)-y2(x-z)+xz(x-z)=0
(x-z)(xy-yz-y2+xz)=0
(x-z)(x-y)(y+z)=0
Nên x-z=0 hoặc x-y=0 hoặc y+z=0
Do đó: x=z hoặc x=y hoặc y=-z
\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Ta có:
\(2\left(2x^2+xy+2y^2\right)=3\left(x^2+y^2\right)+\left(x+y\right)^2\ge\dfrac{3}{2}\left(x+y\right)^2+1\left(x+y\right)^2=\dfrac{5}{2}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Gợi ý. Dùng cái trên.
\(x^2y-xy^2+x^2z-xz^2+y^2z+yz^2=2xyz\)
\(\Leftrightarrow\left(x^2y-xy^2\right)+\left(x^2z-xyz\right)-\left(xz^2-yz^2\right)-\left(xyz-y^2z\right)=0\)
\(\Leftrightarrow xy\left(x-y\right)+xz\left(x-y\right)-z^2\left(x-y\right)-yz\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(xy+xz-z^2-yz\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[x\left(y+z\right)-z\left(y+z\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-z\right)\left(y+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\)\(\left(đpcm\right)\)