Ta có : \(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\frac{\left(\sqrt{5}+2\sqrt[3]{\sqrt{5}-2^{ }}\right)^3}{\sqrt{5}+3-\sqrt{5}}\) 2)³ trong căn bậc nhé mk ko vt đc ( ko bt giải thick thông cảm )

\(=\frac{\sqrt{5}^2-2^2}{3}\)

\(=\frac{1}{3}\)

Vậy \(A=\left(3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2\right)^{2011}=3^{2011}\)

Trả lời

A=(3x³+8x²+2)²⁰¹¹ với x=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}\sqrt{9-6\sqrt{5}+5}}\)

=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\right)^3-3.\left(\sqrt{5}\right)^2.2+3\sqrt{5}.2^2-2^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

=\(\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}\)

=\(\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{3}\)

=1/3

Học tốt !

Ta có

\(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}-\sqrt{14-6\sqrt{5}}}\)

\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3\cdot5\cdot2+3\sqrt{5}\cdot4-8}}{\sqrt{5}-\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}\)

\(=\frac{\sqrt{5}^2-2^2}{3}=\frac{1}{3}\)

Với \(x=\frac{1}{3}\)thay vào bt ta có

\(A=\left[3\cdot\left(\frac{1}{3}\right)^3+8\cdot\left(\frac{1}{3}\right)^2+2\right]^{2011}\)

\(=3^{2011}\)

Sửa đề:

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+12\sqrt{5}-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

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

Thế vô A ta được

\(A=\left(3.\dfrac{1}{3^3}+8.\dfrac{1}{3^2}+2\right)^{2018}=3^{2018}\)

\(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)

\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}\right)^3-3.\left(\sqrt{5}\right)^2.2+3.\sqrt{5}.2^2-2^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}=\frac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{3}\)

\(=\frac{1}{3}\) Chắc được rồi :))

thanks :))

\(x=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\frac{\sqrt[3]{17\sqrt{5}-38}\left(\sqrt{5}+2\right)}{3}=\frac{\sqrt[3]{17\sqrt{5}-38}.\sqrt[3]{\left(\sqrt{5}+2\right)^3}}{3}\)

\(=\frac{\sqrt[3]{\left(17\sqrt{5}-38\right)\left(17\sqrt{5}+38\right)}}{3}=\frac{1}{3}\)

\(\Rightarrow A=\left[3.\left(\frac{1}{3}\right)^3+8.\left(\frac{1}{3}\right)^2+2\right]^{2005}=3^{2005}\)

Mẫu của x

\(\sqrt{5}+\sqrt{3^2-2.3.\sqrt{5}+5}=\sqrt{5}+\left|3-\sqrt{5}\right|=3\)

Tử của x

\(\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}=\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\sqrt{5}\right)-3.\left(\sqrt{5}\right)^2.2+3.\sqrt{5}.2^2-2^3}=\left(\sqrt{5}+2\right)\sqrt{\left(\sqrt{5}-2\right)^3}=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=5-4=1\)

=> \(x=\dfrac{1}{3}\)

\(A=\left(\dfrac{3}{3^3}+\dfrac{8}{3^2}+2\right)^{1998}=\left(\dfrac{1+8+9}{3^2}\right)^{1998}=2^{1998}\)

\(x=\dfrac{\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}.\left(\sqrt{5}+2\right)=\dfrac{\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}.\left(\sqrt{5}+2\right)=\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{5-4}{3}=\dfrac{1}{3}\) Thay : \(x=\dfrac{1}{3}\) vào A , ta được :

\(A=\left(\dfrac{3}{27}+\dfrac{8}{9}-\dfrac{3}{3}+1\right)^{2012}=1^{2012}=1\)

Vậy ,...