Nhang nha cần gấp. Toán 9 rút gọn. Question Nhang nha cần gấp. Toán 9 rút gọn. in progress 0 Môn Toán Khoii Minh 4 years 2020-10-21T07:23:35+00:00 2020-10-21T07:23:35+00:00 2 Answers 73 views 0
Answers ( )
Đáp án:
Giải thích các bước giải:
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Đáp án:
b. \(\dfrac{1}{{\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
\dfrac{{2\sqrt x \left( {\sqrt x – 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) – 3x + 1 – \sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x – 6\sqrt x + x + 4\sqrt x + 3 – 3x + 1 – \sqrt x }}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ – 3\sqrt x + 4}}{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right)}}\\
b.DK:x > 0;x \ne 1\\
\left[ {\dfrac{{x – \sqrt x + 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x – 1 + 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x }}\\
c.DK:a > 0;a \ne 1\\
\left[ {\dfrac{{\sqrt a – \sqrt a + 1}}{{\sqrt a \left( {\sqrt a – 1} \right)}}} \right]:\left( {\dfrac{{\sqrt a + 1 – \sqrt a – 2}}{{\sqrt a – 1}}} \right)\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\sqrt a – 1}}{{ – 1}}\\
= – \dfrac{1}{{\sqrt a }}
\end{array}\)