Tìm x,y,z
\(\dfrac{3x-2y}{37}=\dfrac{5y-3z}{15}=\dfrac{2z-5x}{2}và10x-3y-2z=-4\)
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Giải:
Ta có: \(\dfrac{3x-2y}{5}=\dfrac{5y-3z}{2}=\dfrac{2z-5x}{2}\)
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}=\dfrac{15x-10y+10y-6z+6z-15x}{25+4+6}=0\)
\(\Rightarrow\left\{{}\begin{matrix}15x-10y=0\\10y-6z=0\\6z-15x=0\end{matrix}\right.\Rightarrow15x=10y=6z\)
\(\Rightarrow\dfrac{15x}{30}=\dfrac{10y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=10\\y=15\\z=25\end{matrix}\right.\)
Vậy...
\(\dfrac{3x-2y}{5}=\dfrac{5y-3z}{2}=\dfrac{2z-5x}{2}\)
\(\Rightarrow\dfrac{5\left(3x-2y\right)}{25}=\dfrac{2\left(5y-3z\right)}{4}=\dfrac{3\left(2z-5x\right)}{6}\)
\(\Rightarrow\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{15x-10y}{25}=\dfrac{10y-6z}{4}=\dfrac{6z-15x}{6}\)
\(=\dfrac{15x-10y+10y-6z+6z-15x}{25+4+6}\)
\(=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\5y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{5}\\2z=5x\Rightarrow\dfrac{z}{5}=\dfrac{x}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.5=25\end{matrix}\right.\)
\(\dfrac{3x-2y}{5}\)=\(\dfrac{2z-5x}{3}\)=\(\dfrac{5y-3z}{2}\)
⇒\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)=\(\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}\)=0
⇒3x-2y=2z-5x=5y-3z=0
* 3x-2y=0⇒3x=2y⇒\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)
* 2z-5x=0⇒2z=5x⇒\(\dfrac{z}{5}\)=\(\dfrac{x}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)=\(\dfrac{x+y+z}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5
\(\dfrac{x}{2}\)=-5⇒x=-10
\(\dfrac{y}{3}\)=-5⇒y=-15
\(\dfrac{z}{5}\)=-5⇒z=-25
Vậy x=-10;y=-15;z=-25
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3