Question

giải hệ phương trình:

a)\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\5yz=6\left(y-z\right)\\4xz=3\left(x+y\right)\end{matrix}\right.\)

b)\(\left\{{}\begin{matrix}\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}\\7x-3y+2z=37\end{matrix}\right.\)

Answer 1

a: Sửa đề: 

\(\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\4yz=3\left(y+z\right)\\5xz=6\left(z+x\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{4}{3}\\\dfrac{x+z}{xz}=\dfrac{5}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{4}{3}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{2}\\\dfrac{1}{y}=1\\\dfrac{1}{z}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{2}{3};y=1;z=3\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau,ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{9}=\dfrac{7x-3y+2z}{7\cdot4-3\cdot3+2\cdot9}=\dfrac{37}{37}=1\)

=>x=4; y=3; z=9