Giải hệ phương trình
\(\left\{{}\begin{matrix}6x\left(y^2+z^2\right)=13yz\\3y\left(z^2+x^2\right)=5zx\\6z\left(x^2+y^2\right)=5xy\end{matrix}\right.\)
Giải hệ phương trình tìm nghiệm nguyên:
\(\left\{{}\begin{matrix}\left(2-x\right)\left(3x-2z\right)=3-z\left(1\right)\\y^3+3y=x^2-3x+2\left(2\right)\\z^2+y^2=6z\left(3\right)\\z\le3\left(4\right)\end{matrix}\right.\)
Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}x^3+x\left(y-z\right)^2=2\\y^3+y\left(z-x\right)^2=30\\z^3+z\left(x-y\right)^2=16\end{matrix}\right.\)
Giải hệ phương trình tìm nghiệm nguyên:
\(\left\{{}\begin{matrix}\left(2-x\right)\left(3x-2z\right)=3-z\\y^3+3y=x^2-3x+2\\z^2+y^2=6z\\z\le3\end{matrix}\right.\)
Giải hệ phương trình \(\left\{{}\begin{matrix}x+y^2+z^3=14\\\left(\frac{1}{2x}+\frac{1}{3y}+\frac{1}{6z}\right)\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right)=1\end{matrix}\right.\)
Giải hệ phương trình\(\left\{{}\begin{matrix}\left(x+1\right)\left(x^2+1\right)=y^3+1\\\left(y+1\right)\left(y^2+1\right)=z^3+1\\\left(z+1\right)\left(z^2+1\right)=x^3+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+1\right)\left(x^2+1\right)=y^3+1\\\left(y+1\right)\left(y^2+1\right)=z^3+1\\\left(z+1\right)\left(z^2+1\right)=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+x^2+x=y^3\left(1\right)\\y^3+y^2+y=z^3\\z^3+z^2+z=x^3\end{matrix}\right.\)
Giả sử \(x>y\Rightarrow x^3+x^2+x>y^3+y^2+y\)
\(\Rightarrow y^3>z^3\Leftrightarrow y>z\left(2\right)\)
\(\Rightarrow y^3+y^2+y>z^3+z^2+z\Rightarrow z>x\left(3\right)\)
Từ \(\left(2\right);\left(3\right)\Rightarrow y>x\) (Vô lí)
Giả sử \(x< y\Rightarrow x^3+x^2+x< y^3+y^2+y\)
\(\Rightarrow y^3< z^3\Leftrightarrow y< z\left(4\right)\)
\(\Rightarrow y^3+y^2+y< z^3+z^2+z\Rightarrow z< x\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow y< x\) (Vô lí)
\(\Rightarrow x=y=z\)
\(\left(1\right)\Leftrightarrow x^3+x^2+x=x^3\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow x=y=z=0\) hoặc \(x=y=z=-1\)
Giải các hệ phương trình sau:a) \(\left\{{}\begin{matrix}\left(2x-y\right)^2-6x+3y=0\\x+2y=0\end{matrix}\right.\);b) \(\left\{{}\begin{matrix}\sqrt{\dfrac{2x-y}{x+y}}+\sqrt{\dfrac{x+y}{2x-y}}=2\\3x+y=14\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
Giai hệ phương trình:
\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5zx=6\left(z+x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5zx=6\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{z+x}{zx}=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\)
Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\b+c=\dfrac{4}{3}\\a+c=\dfrac{5}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=1\\c=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\\z=3\end{matrix}\right.\)
Vậy . . .
Giải hệ phương trình: \(\left\{{}\begin{matrix}x^3\left(y^2+3y+3\right)=3y^2\\y^3\left(z^2+3z+3\right)=3z^2\\z^3\left(x^2+3x+3\right)=3x^2\end{matrix}\right.\)
giải hệ phương trình bằng pp sd bđt:
\(\left\{{}\begin{matrix}x+y^2+z^3=14\\\left(\dfrac{1}{2x}+\dfrac{1}{3y}+\dfrac{1}{6z}\right)\left(\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{6}\right)=1\end{matrix}\right.\)