Question

Answer 1

\(a,HPT\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{3}x-\dfrac{2}{3}y=3\left(1\right)\\\dfrac{2}{3}y=x-3\left(2\right)\end{matrix}\right.\)

Thế \(\left(2\right)\) vào \(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-x+3=3\Leftrightarrow x=0\Leftrightarrow\dfrac{2}{3}y=-3\Leftrightarrow y=-\dfrac{9}{2}\)

\(b,HPT\Leftrightarrow\left\{{}\begin{matrix}\sqrt{5}x-y=2+2y\left(1\right)\\\sqrt{2}\left(\sqrt{5}x-y\right)=1\left(2\right)\end{matrix}\right.\)

Thế \(\left(1\right)\) vào \(\left(2\right)\Leftrightarrow\sqrt{2}\left(2+2y\right)=1\Leftrightarrow2+2y=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow2y=\dfrac{\sqrt{2}-4}{2}\Leftrightarrow y=\dfrac{\sqrt{2}-4}{4}\\ \Leftrightarrow\sqrt{5}x=2+3y=\dfrac{3\sqrt{2}-4}{4}\\ \Leftrightarrow x=\dfrac{3\sqrt{2}-4}{4\sqrt{5}}=\dfrac{3\sqrt{10}-4\sqrt{5}}{20}\)

Answer 2

a) \(hpt\Leftrightarrow\left\{{}\begin{matrix}4\left(3+\dfrac{2}{3}y\right)-2y=9\\x=3+\dfrac{2}{3}y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3+\dfrac{2}{3}y\\\dfrac{2}{3}y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{9}{2}\end{matrix}\right.\)

b) \(hpt\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+3y}{\sqrt{5}}\\\sqrt{10}.\dfrac{2+3y}{\sqrt{5}}-\sqrt{2}y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+3y}{\sqrt{5}}\\2\sqrt{2}y=1-2\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\sqrt{10}-4\sqrt{5}}{20}\\y=\dfrac{-4+\sqrt{2}}{4}\end{matrix}\right.\)

Answer 3

a) \(\left\{{}\begin{matrix}4x-2y=9\left(1\right)\\x-\dfrac{2}{3}y=3\left(2\right)\end{matrix}\right.\)

Từ (2) \(\Leftrightarrow x=3+\dfrac{2}{3}y\)  (3) thế vào (1), ta có:

\(4.\left(3+\dfrac{2}{3}y\right)-2y=9\)

\(\Leftrightarrow12+\dfrac{8}{3}y-2y=9\)

\(\Leftrightarrow\dfrac{2}{3}y=-3\)

\(\Leftrightarrow y=\dfrac{-9}{2}\) thế vào (3), ta có:

\(x=3+\dfrac{2}{3}.\left(-\dfrac{9}{2}\right)\)

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{\left(0;-\dfrac{9}{2}\right)\right\}\)

Answer 4

b) \(\left\{{}\begin{matrix}\sqrt{5}x-3y=2\left(1\right)\\\sqrt{10}x-\sqrt{2}y=1\left(2\right)\end{matrix}\right.\)

Từ (1) \(\Leftrightarrow y=\dfrac{\sqrt{5}x-2}{3}\) (3) thế vào (2), ta có:

\(\sqrt{10}x-\sqrt{2}.\left(\dfrac{\sqrt{5}x-2}{3}\right)=1\)

\(\Leftrightarrow\sqrt{10}x-\dfrac{\sqrt{10}x}{3}+\dfrac{2\sqrt{2}}{3}=1\)

\(\Leftrightarrow\dfrac{2\sqrt{10}x}{3}=\dfrac{3-2\sqrt{2}}{3}\)

\(\Leftrightarrow x=\dfrac{3-2\sqrt{2}}{2\sqrt{10}}=\dfrac{3\sqrt{10}-4\sqrt{5}}{20}\) thế vào (3) ta có:

\(y=\dfrac{\sqrt{5}\left(\dfrac{3\sqrt{10}-4\sqrt{5}}{20}\right)-2}{3}=\dfrac{15\sqrt{2}-60}{60}=\dfrac{1}{2\sqrt{2}}-1\)

Vậy \(S=\left\{\left(\dfrac{3\sqrt{10}-4\sqrt{5}}{20};\dfrac{1}{2\sqrt{2}}-1\right)\right\}\)