Z=(X+2/√X+1 – √x ):(√x – 4/ 1-x – √x/ √x +1 03/10/2021 Bởi Brielle Z=(X+2/√X+1 – √x ):(√x – 4/ 1-x – √x/ √x +1
$$\eqalign{ & \left( {{{x + 2} \over {\sqrt x + 1}} – \sqrt x } \right):\left( {{{\sqrt x – 4} \over {1 – x}} – {{\sqrt x } \over {\sqrt x + 1}}} \right) \cr & = {{x + 2 – x – \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x – 4 – \sqrt x \left( {1 – \sqrt x } \right)} \over {\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr & = {{2 – \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x – 4 – \sqrt x + x} \over {\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr & = {{2 – \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {x – 4}} \cr & = {{2 – \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}} \cr & = {{\sqrt x – 1} \over {\sqrt x + 2}} \cr} $$ Bình luận
\[\begin{array}{l} Z = \left( {\frac{{x + 2}}{{\sqrt x + 1}} – \sqrt x } \right):\left( {\frac{{\sqrt x – 4}}{{1 – x}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\,\,\,\left( {DK:\,\,x \ge 0;\,\,x \ne 1} \right)\\ = \frac{{x + 2 – x – \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x – 4 – \sqrt x \left( {1 – \sqrt x } \right)}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x – 4 – \sqrt x + x}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{x – 4}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\ = \frac{{2 – \sqrt x }}{{\sqrt x + 1}}.\frac{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\ = \frac{{1 – \sqrt x }}{{ – \left( {\sqrt x + 2} \right)}} = \frac{{\sqrt x – 1}}{{\sqrt x + 2}}. \end{array}\] Bình luận
$$\eqalign{
& \left( {{{x + 2} \over {\sqrt x + 1}} – \sqrt x } \right):\left( {{{\sqrt x – 4} \over {1 – x}} – {{\sqrt x } \over {\sqrt x + 1}}} \right) \cr
& = {{x + 2 – x – \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x – 4 – \sqrt x \left( {1 – \sqrt x } \right)} \over {\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr
& = {{2 – \sqrt x } \over {\sqrt x + 1}}:{{\sqrt x – 4 – \sqrt x + x} \over {\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}} \cr
& = {{2 – \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {x – 4}} \cr
& = {{2 – \sqrt x } \over {\sqrt x + 1}}.{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)} \over {\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}} \cr
& = {{\sqrt x – 1} \over {\sqrt x + 2}} \cr} $$
\[\begin{array}{l}
Z = \left( {\frac{{x + 2}}{{\sqrt x + 1}} – \sqrt x } \right):\left( {\frac{{\sqrt x – 4}}{{1 – x}} – \frac{{\sqrt x }}{{\sqrt x + 1}}} \right)\,\,\,\left( {DK:\,\,x \ge 0;\,\,x \ne 1} \right)\\
= \frac{{x + 2 – x – \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x – 4 – \sqrt x \left( {1 – \sqrt x } \right)}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{\sqrt x – 4 – \sqrt x + x}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \frac{{2 – \sqrt x }}{{\sqrt x + 1}}:\frac{{x – 4}}{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \frac{{2 – \sqrt x }}{{\sqrt x + 1}}.\frac{{\left( {1 – \sqrt x } \right)\left( {1 + \sqrt x } \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \frac{{1 – \sqrt x }}{{ – \left( {\sqrt x + 2} \right)}} = \frac{{\sqrt x – 1}}{{\sqrt x + 2}}.
\end{array}\]