Giải thích các bước giải:
Câu 1:
a. $\left \{ {{3x+y=1} \atop {-x-2y=-3}} \right.$
<-> $\left \{ {{y=1-3x} \atop {-x-2(1-3x)=-3}} \right.$
<-> $\left \{ {{y=1-3x} \atop {-x-2+6x=-3}} \right.$
<-> $\left \{ {{y=1-3x} \atop {x=\frac{-1}{5}}} \right.$
<-> $\left \{ {{y=\frac{8}{5}} \atop {x=\frac{-1}{5}}} \right.$
b. $\sqrt[]{2x+3}$ =2-x
<-> $\left \{ {{2-x>0} \atop {2x+3=4-4x+x^2}} \right.$
<-> $\left \{ {{x<2} \atop {x^2-6x+1=0}} \right.$
<-> $\left \{ {{x<2} \atop {\left[ \begin{array}{l}x=3+2√2\\x=3-2√2\end{array} \right. }} \right.$ <-> x=3-2√2
Câu 2:
a. Vì MNPQ là hình bình hành
\(\begin{array}{l}
\to \overrightarrow {MN} = \overrightarrow {QP} \\
\leftrightarrow \left\{ \begin{array}{l}
3 - 0 = - 2 - x\\
5 - 3 = 1 - y
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x = - 5\\
y = - 1
\end{array} \right. \to Q( - 5, - 1)
\end{array}\)
\(\begin{array}{l}
b.\overrightarrow {MN} = (3,2) \to MN = \sqrt {{3^2} + {2^2}} = \sqrt {13} \\
\overrightarrow {MP} = ( - 2, - 2) \to MP = \sqrt {{{( - 2)}^2} + {{( - 2)}^2}} = 2\sqrt 2 \\
\overrightarrow {NP} = ( - 5, - 4) \to NP = \sqrt {{{( - 5)}^2} + {{( - 4)}^2}} = \sqrt {41} \\
\to {C_{MNP}} = MN + MP + NP = \sqrt {13} + 2\sqrt 2 + \sqrt {41}
\end{array}\)
