Đáp án:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a\ S=\{1;4\}\\ b.\ ( x;y) =\left( -\frac{4}{7} ;\frac{36}{7}\right)\\ c.\ 2\sqrt{x} -1\\ d.\ S=\{4\} \end{array}$
Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} a\ ĐK:\ x\geqslant 0\\ \Leftrightarrow \left( 1-\sqrt{x}\right)\left(\sqrt{x} -2\right) =0\\ \Leftrightarrow \sqrt{x} =1;\ \sqrt{x} =2\\ \Leftrightarrow x=1\ ( TM) ;\ x=4\ ( TM)\\ Vậy\ S=\{1;4\}\\ b.\ \Leftrightarrow \{_{3x-2y=-12}^{y=4-2x} \Leftrightarrow \{_{3x-2( 4-2x) =-12}^{y=4-2x}\\ \Leftrightarrow \{_{x=-\frac{4}{7}}^{y=4-2x} \Leftrightarrow \{_{x=-\frac{4}{7}}^{y=\frac{36}{7}}\\ Vậy\ ( x;y) =\left( -\frac{4}{7} ;\frac{36}{7}\right)\\ c.\ =\frac{\sqrt{x}\left(\sqrt{x} +1\right)}{\sqrt{x}} +\frac{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)}{\sqrt{x} +2}\\ =\sqrt{x} +1+\sqrt{x} -2=2\sqrt{x} -1\\ d.\ ĐK:\ -\frac{1}{2} \leqslant x\leqslant 7\\ \Leftrightarrow 2x+1=x^{2} +49-14x\\ \Leftrightarrow x^{2} -16x+48=0\\ \Leftrightarrow ( x-12)( x-4) =0\\ \Leftrightarrow x=12\ ( loại) ;\ x=4\ ( TM)\\ Vậy\ S=\{4\} \end{array}$
