Question

Cine mă ajută și pe mine?
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Answer 1

$\it \begin{cases}\it 3\sqrt2x-y=7\\ \\ \it 2\sqrt2x+y=3\end{cases}\\ \rule{90}{0.8}\\ \rule{0.2}{0.2}\ \ \ 5\sqrt2x \ \ \ \ \ =10\\ \\ 5\sqrt2x=10\bigg|_{:5} \Rightarrow \sqrt2x=2 \Rightarrow \sqrt2x=\sqrt2\cdot\sqrt2\bigg|_{:\sqrt2} \Rightarrow x=\sqrt2$

Înlocuim în a doua ecuație a sistemului valoarea lui x :

$\it 2\sqrt2\cdot\sqrt2+y=3 \Rightarrow 4+y=3\Rightarrow y=3-4\Rightarrow y=-1\\ \\ S=\{(\sqrt2,\ \ -1)\}$

Answer 2

Rezolvare prin metoda reducerii:

$\left\big \{\begin{matrix} 3\sqrt{2}x - y = 7\\ 2\sqrt{2}x + y = 3 \end{matrix}\right.$

$3\sqrt{2} x + 2\sqrt{2} x = 7 + 3 \iff 5\sqrt{2} x = 10 \ \Big|:(5\sqrt{2})$

$x = \dfrac{10^{(\sqrt{2} } }{5\sqrt{2}} = \dfrac{10\sqrt{2}}{10} \implies \bf x = \sqrt{2}$

$3\sqrt{2} \cdot \sqrt{2} - y = 7 \iff 6 - y = 7 \iff y = 6 - 7$

$\implies \bf y = -1$

$\implies S =\Big\{(\sqrt{2} ;-1)\Big\}$

Rezolvare prin metoda substituției:

$\left\big \{\begin{matrix} 3\sqrt{2}x - y = 7\\ 2\sqrt{2}x + y = 3 \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} 3\sqrt{2}x - y = 7\\ y = 3 - 2\sqrt{2}x \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} 3\sqrt{2}x - (3 - 2\sqrt{2}x) = 7\\ y = 3 - 2\sqrt{2}x \end{matrix}\right.$

$\left\big \{\begin{matrix} 3\sqrt{2}x - 3 + 2\sqrt{2}x) = 7\\ y = 3 - 2\sqrt{2}x \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} 5\sqrt{2}x = 7 + 3\\ y = 3 - 2\sqrt{2}x \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} 5\sqrt{2}x = 10\\ y = 3 - 2\sqrt{2}x \end{matrix}\right.$

$\left\big \{\begin{matrix} x = \dfrac{10^{(\sqrt{2} } }{5\sqrt{2}} \\ y = 3 - 2\sqrt{2}x \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} x = \dfrac{10\sqrt{2}}{10} \\ y = 3 - 2\sqrt{2}x \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} x = \sqrt{2} \\ y = 3 - 2\sqrt{2}x \end{matrix}\right.$

$\left\big \{\begin{matrix} x = \sqrt{2} \\ y = 3 - 2\sqrt{2} \cdot \sqrt{2} \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} x = \sqrt{2} \\ y = 3 - 4 \end{matrix}\right. \Leftrightarrow \left\big \{\begin{matrix} x = \sqrt{2} \\ y = -1\end{matrix}\right.$

$\implies S =\Big\{(\sqrt{2} ;-1)\Big\}$