Rút gọn : A = 2x-√x /√x -x + x /√x -1 B = ( √x +2 / x+2√x +1 – √x -2 /x-1) . √x +1 / √x 04/08/2021 Bởi Elliana Rút gọn : A = 2x-√x /√x -x + x /√x -1 B = ( √x +2 / x+2√x +1 – √x -2 /x-1) . √x +1 / √x
Đáp án: Giải thích các bước giải: A=2x−√x√x−x+x√x−1(x>0,x≠1)=√x(2√x−1)√x(1−√x)+x√x−1=2√x−11−√x−x1−√x=2√x−1−x1−√x=x−2√x+1√x−1=(√x−1)2√x−1=√x−1B=(√x+2x+2√x+1−√x−2x−1).√x+1√x(x>0,x≠1)=⎛⎝√x+2(√x+1)2−√x−2(√x−1)(√x+1)⎞⎠.√x+1√x=(√x+2)(√x−1)−(√x−2)(√x+1)(√x+1)2(√x−1).√x+1√x=(x+√x−2)−(x−√x−2)(√x+1).(√x−1).1√x=2√x(√x+1)(√x−1).1√x=2x−1 Bình luận
Giải thích các bước giải: Ta có: \(\begin{array}{l}A = \dfrac{{2x – \sqrt x }}{{\sqrt x – x}} + \dfrac{x}{{\sqrt x – 1}}\,\,\,\,\,\,\,\,\,\left( {x > 0,\,\,x \ne 1} \right)\\ = \dfrac{{\sqrt x \left( {2\sqrt x – 1} \right)}}{{\sqrt x \left( {1 – \sqrt x } \right)}} + \dfrac{x}{{\sqrt x – 1}}\\ = \dfrac{{2\sqrt x – 1}}{{1 – \sqrt x }} – \dfrac{x}{{1 – \sqrt x }}\\ = \dfrac{{2\sqrt x – 1 – x}}{{1 – \sqrt x }}\\ = \dfrac{{x – 2\sqrt x + 1}}{{\sqrt x – 1}}\\ = \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\sqrt x – 1}}\\ = \sqrt x – 1\\B = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} – \dfrac{{\sqrt x – 2}}{{x – 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\,\,\,\,\,\,\,\,\left( {x > 0,x \ne 1} \right)\\ = \left( {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} – \dfrac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\ = \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right) – \left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x – 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\ = \dfrac{{\left( {x + \sqrt x – 2} \right) – \left( {x – \sqrt x – 2} \right)}}{{\left( {\sqrt x + 1} \right).\left( {\sqrt x – 1} \right)}}.\dfrac{1}{{\sqrt x }}\\ = \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}.\dfrac{1}{{\sqrt x }}\\ = \dfrac{2}{{x – 1}}\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
A=2x−√x√x−x+x√x−1(x>0,x≠1)=√x(2√x−1)√x(1−√x)+x√x−1=2√x−11−√x−x1−√x=2√x−1−x1−√x=x−2√x+1√x−1=(√x−1)2√x−1=√x−1B=(√x+2x+2√x+1−√x−2x−1).√x+1√x(x>0,x≠1)=⎛⎝√x+2(√x+1)2−√x−2(√x−1)(√x+1)⎞⎠.√x+1√x=(√x+2)(√x−1)−(√x−2)(√x+1)(√x+1)2(√x−1).√x+1√x=(x+√x−2)−(x−√x−2)(√x+1).(√x−1).1√x=2√x(√x+1)(√x−1).1√x=2x−1
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{2x – \sqrt x }}{{\sqrt x – x}} + \dfrac{x}{{\sqrt x – 1}}\,\,\,\,\,\,\,\,\,\left( {x > 0,\,\,x \ne 1} \right)\\
= \dfrac{{\sqrt x \left( {2\sqrt x – 1} \right)}}{{\sqrt x \left( {1 – \sqrt x } \right)}} + \dfrac{x}{{\sqrt x – 1}}\\
= \dfrac{{2\sqrt x – 1}}{{1 – \sqrt x }} – \dfrac{x}{{1 – \sqrt x }}\\
= \dfrac{{2\sqrt x – 1 – x}}{{1 – \sqrt x }}\\
= \dfrac{{x – 2\sqrt x + 1}}{{\sqrt x – 1}}\\
= \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\sqrt x – 1}}\\
= \sqrt x – 1\\
B = \left( {\dfrac{{\sqrt x + 2}}{{x + 2\sqrt x + 1}} – \dfrac{{\sqrt x – 2}}{{x – 1}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\,\,\,\,\,\,\,\,\left( {x > 0,x \ne 1} \right)\\
= \left( {\dfrac{{\sqrt x + 2}}{{{{\left( {\sqrt x + 1} \right)}^2}}} – \dfrac{{\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 1} \right) – \left( {\sqrt x – 2} \right)\left( {\sqrt x + 1} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}\left( {\sqrt x – 1} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {x + \sqrt x – 2} \right) – \left( {x – \sqrt x – 2} \right)}}{{\left( {\sqrt x + 1} \right).\left( {\sqrt x – 1} \right)}}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{{2\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}.\dfrac{1}{{\sqrt x }}\\
= \dfrac{2}{{x – 1}}
\end{array}\)