x _ 1 + 1 x-3 2- can x can x + 2 rut gon ho minh 02/08/2021 Bởi Alice x _ 1 + 1 x-3 2- can x can x + 2 rut gon ho minh
Đáp án: \(\dfrac{{{x^2} – 4x + 2x\sqrt x – 6\sqrt x }}{{\left( {x – 3} \right)\left( {x – 4} \right)}}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ge 0;x \ne 3;x \ne 4\\\dfrac{x}{{x – 3}} – \dfrac{1}{{2 – \sqrt x }} + \dfrac{1}{{\sqrt x + 2}}\\ = \dfrac{x}{{x – 3}} + \dfrac{{\sqrt x + 2 + \sqrt x – 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}\\ = \dfrac{x}{{x – 3}} + \dfrac{{2\sqrt x }}{{x – 4}}\\ = \dfrac{{{x^2} – 4x + 2x\sqrt x – 6\sqrt x }}{{\left( {x – 3} \right)\left( {x – 4} \right)}}\end{array}\) Bình luận
Đáp án:
\(\dfrac{{{x^2} – 4x + 2x\sqrt x – 6\sqrt x }}{{\left( {x – 3} \right)\left( {x – 4} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 3;x \ne 4\\
\dfrac{x}{{x – 3}} – \dfrac{1}{{2 – \sqrt x }} + \dfrac{1}{{\sqrt x + 2}}\\
= \dfrac{x}{{x – 3}} + \dfrac{{\sqrt x + 2 + \sqrt x – 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{x}{{x – 3}} + \dfrac{{2\sqrt x }}{{x – 4}}\\
= \dfrac{{{x^2} – 4x + 2x\sqrt x – 6\sqrt x }}{{\left( {x – 3} \right)\left( {x – 4} \right)}}
\end{array}\)