Question

\(\left\{{}\begin{matrix}x-\sqrt{3}y=o\\\\\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\)

giải hệ phương trình sau bằng phương pháp thế ;giúp mình với nhé

Answer 1

hpt tương đương \(\left\{{}\begin{matrix}x=\sqrt{3}y\\\sqrt{3}x+2y=1+\sqrt{3}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\sqrt{3}y\\\sqrt{3}.\sqrt{3}y+2y=1+\sqrt{3}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\sqrt{3}y\\3y+2y=1+\sqrt{3}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\sqrt{3}y\\5y=1+\sqrt{3}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\sqrt{3}.\dfrac{1+\sqrt{3}}{5}=\dfrac{\sqrt{3}+3}{5}\\y=\dfrac{1+\sqrt{3}}{5}\end{matrix}\right.\)

Answer 2

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}y\\3y+2y=1+\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{3}y\\y=\dfrac{1+\sqrt{3}}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3+\sqrt{3}}{5}\\y=\dfrac{1+\sqrt{3}}{5}\end{matrix}\right.\)

Vậy..............