\(a>0\)

Có \(a^3=2-\sqrt{3}+3\sqrt[3]{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\left(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}\right)+2+\sqrt{3}\)

\(\Leftrightarrow a^3=4+3a\) 

\(\Leftrightarrow a\left(a^2-3\right)=4\)\(\Leftrightarrow a^2-3=\dfrac{4}{a}\)

\(\Leftrightarrow\dfrac{64}{\left(a^2-3\right)^3}=a^{.3}\) 

\(\Leftrightarrow\dfrac{64}{\left(a^2-3\right)^3}-3a=a^2-3a=4\) là số nguyên.

Ta có : a= \(\sqrt[3]{2-\sqrt{3}}\)  + \(\sqrt[3]{2+\sqrt{3}}\)

Suy ra a^3 = 3a +4  => (a^2 -3)a=4  

<=> \(\left(\frac{4}{a^2-3}\right)^3\)= a^3  <=>\(\frac{64}{\left(a^2-a\right)^3}\) -3a = 4   

mà 4 nguyên suy ra đpcm

Ta có \(a=3\sqrt{2-\sqrt{3}}+\sqrt{3}^32_{\sqrt{3}}\)

Suy ra ta được 3= 3a + 4 => (a ngũ 2 - 3)a =4

Vậy kết quả khi tính đ là

=> (4 trên a2 - 3) trên 3 =a ngũ 3 <=> 64 trên a 2 - a3 - 3a =4

\(A=\left(\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right)\left(\dfrac{\sqrt{3}-1}{3\sqrt{2}-\sqrt{6}}\right)\)

\(=\dfrac{5+2\sqrt{6}-5+2\sqrt{6}}{-1}\cdot\dfrac{1}{\sqrt{6}}\)

=-4

=5+2√6−5+2√6−1⋅1√6=5+26−5+26−1⋅16

=-4

a: Sửa đề: căn 6+2căn 5-căn 5

\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)

b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)

=>a^3-3a-4=0

=>a^3-3a=4

\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)

\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)

=4

Bạn tham khảo câu số 9:

mọi người giúp em mấy bài này với ạ =((( - Hoc24

Xem lại câu hỏi

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)