A=(1-√x/√x+1):(√x+3/
√x-2+√x+2/3-√x+√x+2/x-5√x+6)
Rút gọn bieu thu A
0 bình luận về “A=(1-√x/√x+1):(√x+3/
√x-2+√x+2/3-√x+√x+2/x-5√x+6)
Rút gọn bieu thu A”
Đáp án:
$\begin{array}{l} Dkxd:x \ge 0;x \ne 1;x \ne 4;x \ne 9\\ A = \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}:\left( {\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} + \dfrac{{\sqrt x + 2}}{{3 – \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}}} \right)\\ = \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}:\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{x – 9 – x + 4 + \sqrt x + 2}}\\ = \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{\sqrt x – 1}}\\ = \dfrac{{ – \left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{\sqrt x + 1}}\\ = \dfrac{{ – x + 5\sqrt x – 6}}{{\sqrt x + 1}} \end{array}$
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1;x \ne 4;x \ne 9\\
A = \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}:\left( {\dfrac{{\sqrt x + 3}}{{\sqrt x – 2}} + \dfrac{{\sqrt x + 2}}{{3 – \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x – 5\sqrt x + 6}}} \right)\\
= \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}:\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x – 3} \right) – \left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{x – 9 – x + 4 + \sqrt x + 2}}\\
= \dfrac{{1 – \sqrt x }}{{\sqrt x + 1}}.\dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{\sqrt x – 1}}\\
= \dfrac{{ – \left( {\sqrt x – 3} \right)\left( {\sqrt x – 2} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{ – x + 5\sqrt x – 6}}{{\sqrt x + 1}}
\end{array}$