Toán Lớp 9: √ a + 1/ a√a + a + √a . ( a^2 – √ a ) Rút gọn
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Toán Lớp 9: √ a + 1/ a√a + a + √a . ( a^2 – √ a ) Rút gọn
Comments ( 1 )
DK:a > 0\\
\dfrac{{\sqrt a + 1}}{{a\sqrt a + a + \sqrt a }}.\left( {{a^2} – \sqrt a } \right)\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {a + \sqrt a + 1} \right)}}.\sqrt a \left( {a\sqrt a – 1} \right)\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a \left( {a + \sqrt a + 1} \right)}}.\sqrt a \left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)\\
= \left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right)\\
= a – 1
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