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mẫu rút gọn như sau:
\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
xong cộng với cái ở ngoài lại ra 4+2 căn 3 làm tương tự
a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)
b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)
c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)
d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)
\(26n^3-\left(n+2\right)^3-\left(n-2\right)^3=24n^3-24n=24n\left(n-1\right)\left(n+1\right)\)
\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{\left(n+1\right)-\left(n-1\right)}{2\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{2\left(n-1\right)n}-\dfrac{1}{2n\left(n+1\right)}\)
Do đó:
\(VT=\dfrac{1}{24}\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2019.2020.2021}\right)\)
\(=\dfrac{1}{48}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2019.2020}-\dfrac{1}{2020.2021}\right)\)
\(=\dfrac{1}{48}\left(\dfrac{1}{2}-\dfrac{1}{2020.2021}\right)< \dfrac{1}{48}.\dfrac{1}{2}=\dfrac{1}{96}\)
a) \(\sqrt[3]{27a^3}\) - 2a
= 3a - 2a = a
b) \(\sqrt[3]{27a^3}\) - \(\sqrt[3]{-8a^3}\)- \(\sqrt[3]{125a^3}\)
= 3a + 2a - 5a = 0
???? đừng làm thế
what is this ????