Toán Lớp 9: (a+√a/√a+1 + a -4√a +4/√a+2)×3/√a-1
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Toán Lớp 9: (a+√a/√a+1 + a -4√a +4/√a+2)×3/√a-1
Giúp mk với
Comments ( 1 )
DK:x \ge 0;a \ne 1\\
A = \left( {\dfrac{{a + \sqrt a }}{{\sqrt a + 1}} + \dfrac{{a – 4\sqrt a + 4}}{{\sqrt a + 2}}} \right).\dfrac{3}{{\sqrt a – 1}}\\
= \left[ {\dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}} + \dfrac{{a – 4\sqrt a + 4}}{{\sqrt a + 2}}} \right].\dfrac{3}{{\sqrt a – 1}}\\
= \left( {\sqrt a + \dfrac{{a – 4\sqrt a + 4}}{{\sqrt a + 2}}} \right).\dfrac{3}{{\sqrt a – 1}}\\
= \dfrac{{a + 2\sqrt a + a – 4\sqrt a + 4}}{{\sqrt a + 2}}.\dfrac{3}{{\sqrt a – 1}}\\
= \dfrac{{2a – 2\sqrt a + 4}}{{\sqrt a + 2}}.\dfrac{3}{{\sqrt a – 1}}\\
= \dfrac{{6a – 6\sqrt a + 12}}{{a + \sqrt a – 2}}
\end{array}\)