a: \(\left\{{}\begin{matrix}7x-3y=5\\4x+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-3y=5\\y=2-4x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-2\left(2-4x\right)=5\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-4+8x=5\\y=2-4x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}15x=9\\y=2-4x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{5}\\y=2-4\cdot\dfrac{3}{5}=2-\dfrac{12}{5}=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}x\sqrt{5}-y=\sqrt{5}\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}15x-3\sqrt{5}\cdot y=15\left(\sqrt{3}-1\right)\\2\sqrt{3}x+3\sqrt{5}y=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\cdot15-3\sqrt{5}\cdot y+2\sqrt{3}\cdot x+3\sqrt{5}y=15\sqrt{3}-15+21\\x\sqrt[]{5}-y=\sqrt{5}\left(\sqrt{3}-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(15+2\sqrt{3}\right)=15\sqrt{3}+6\\y=x\sqrt{5}-\sqrt{5}\left(\sqrt{3}-1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{15\sqrt{3}+6}{15+2\sqrt{3}}=\sqrt{3}\\y=\sqrt{3}\cdot\sqrt{5}-\sqrt{5}\cdot\sqrt{3}+\sqrt{5}=\sqrt{5}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}1,7x-2y=3,8\\2,1x+5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8,5x-10y=19\\4,2x+10y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}8,5x-10y+4,2x+10y=19+0,8\\1,7x-2y=3,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}12,7x=19,8\\2y=1,7x-3,8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{198}{127}\\y=\dfrac{1,7x-3,8}{2}=-\dfrac{73}{127}\end{matrix}\right.\)
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Chứng minh 2 ý đó giúp e vs ạ e đang cần gấp