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ĐK:: x,y,z\(\ne0\)
\(\left\{{}\begin{matrix}x+y+z=9\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\\xy+yz+zx=27\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=9\\xy+yz+zx=xyz\\xy+xz+yz=27\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+y+z=9\\xyz=27\\xy+yz+xz=27\end{matrix}\right.\)
Coi x;y;z là ba nghiệm x1;x2;x3 của một phương trình bậc ba. Theo công thức Vi-ét ta có: \(\left\{{}\begin{matrix}x_1+x_2+x_3=9\\x_1x_2+x_2x_3+x_3x_1=27\\x_1x_2x_3=27\end{matrix}\right.\)
Suy ra x1;x2;x3 là ba nghiệm của phương trình
\(X^3-9X^2+27X-27=0\Leftrightarrow\left(X-3\right)^3=0\Leftrightarrow X=3\)
Vậy (x;y;z)=(3;3;3)
a: \(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+4z=8\\2x-y+3z=6\\2x-6y+8z=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y+z=2\\8y-4z=1\\x+y+2z=4\end{matrix}\right.\)
=>y=9/20; z=13/20; x=4-y-2z=9/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}z=23-x-y\\z=31-y-t\\z=27-t-x\\x+y+t=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-y+23=-y-t+31\\-y-t-31=-x-t+27\\x+y+t=33\\z=23-x-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+t=8\\x-y=58\\x+y+t=33\\z=23-x-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=x+8\\y=x-58\\x-58+x+8+x=33\\z=23-x-y\end{matrix}\right.\)
=>x=83/3; t=107/3; y=-91/3; z=23-83/3+91/3=77/3
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix}
\frac{1}{x}+\frac{1}{y}=\frac{3}{8}\\
\frac{1}{y}+\frac{1}{z}=\frac{3}{4}\\
\frac{1}{z}+\frac{1}{x}=\frac{5}{6}\end{matrix}\right.\Rightarrow 2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{8}+\frac{3}{4}+\frac{5}{6}\)
\(\Leftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{47}{48}\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z}=\frac{47}{48}-\frac{3}{8}\\ \frac{1}{x}=\frac{47}{48}-\frac{3}{4}\\ \frac{1}{y}=\frac{47}{48}-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{48}{29}\\ y=\frac{48}{11}\\ z=\frac{48}{7}\end{matrix}\right.\)
b) Áp dụng bđt Svac-xơ:
\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)
=> hpt vô nghiệm
c) Ở đây x,y,z là các số thực dương
Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)
\(\left\{{}\begin{matrix}\dfrac{xy}{x+y}=\dfrac{12}{5}\\\dfrac{yz}{y+z}=\dfrac{18}{5}\\\dfrac{zx}{z+x}=\dfrac{36}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{5}{12}\\\dfrac{y+z}{yz}=\dfrac{5}{18}\\\dfrac{z+x}{zx}=\dfrac{13}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{12}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{18}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{13}{36}\end{matrix}\right.\)
Cộng vế theo vế ta thu được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{19}{18}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{19}{36}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{4}\\\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{z}=\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=9\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(4;6;9\right)\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{1}{2}\\\dfrac{y+z}{yz}=\dfrac{1}{4}\\\dfrac{z+x}{xz}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{4}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\) ( đk : x , y , z # 0 )
Cộng từng vế của các pt lại với nhau , ta có :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{12}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{13}{24}-\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{13}{24}-\dfrac{1}{4}=\dfrac{7}{24}\)
\(\Leftrightarrow x=\dfrac{24}{7}\left(tm\right)\)
\(\Rightarrow y=\dfrac{24}{5}\left(tm\right);z=8\left(tm\right)\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=\dfrac{9}{9}=1\)
Dau bang xay ra khi x=y=z=3 ( vi x+y+z=9)