Em tách từng câu ra để hỏi thôi nhé.
Giải thích các bước giải:
\(\begin{array}{l}
2)a)\,DK:\,x \ne 1;x > 0\\
b)A = \dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}} = \dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}} = \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
c)\,A = \dfrac{1}{2}\\
\Rightarrow \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{1}{2} \Rightarrow 2\sqrt x - 6 = \sqrt x - 2\\
\Leftrightarrow \sqrt x = 4 \Leftrightarrow x = 16\left( {tm} \right)
\end{array}\)
\(\begin{array}{l}
3)\,a)\,do\,thi\,ham\,so\,di\,\,qua\,diem\,co\,toa\,do\,\left( {0;1} \right)\\
\Rightarrow 1 = \dfrac{1}{2}.0 + b \Rightarrow b = 1 \Rightarrow y = \dfrac{1}{2}x + 1\\
b)\,HS\,tu\,ve\,hinh\\
c)\,giao\,diem\,voi\,truc\,tung:\,B\left( {0;1} \right) \Rightarrow OB = 1\\
Giao\,diem\,voi\,truc\,hoanh:\,A\left( { - 2;0} \right) \Rightarrow OA = 2\\
OH \bot AB\\
\Rightarrow \dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} = \dfrac{1}{{{1^2}}} + \dfrac{1}{{{2^2}}} = \dfrac{5}{4} \Rightarrow OH = \dfrac{{2\sqrt 5 }}{5}
\end{array}\)
