a = \(\sqrt[3]{26+15\sqrt{3}}\)+\(\sqrt[3]{26-15\sqrt{3}}\)=\(\sqrt[3]{8+2.3.3+3.4.\sqrt{3}+3\sqrt{3}}+\sqrt[3]{8-3.4.\sqrt{3}+2.3.3-3\sqrt{3}}\)

=\(\sqrt[3]{2+\sqrt{3}}^3\)+\(\sqrt[3]{2-\sqrt{3}}^3\)

=2+\(\sqrt{3}\)+2-\(\sqrt{3}\)

=4=\(2^2\)

Ta có \(a=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}+\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}=\sqrt[3]{2^3+3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}+\sqrt[3]{2^3-3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2-\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{\left(2-\sqrt{3}\right)^3}=2+\sqrt{3}+2-\sqrt{3}=4=2^2\)

Vậy a là bình phương của một số nguyên

Đặt \(x=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt[]{3}}\)

\(\Rightarrow x^3=52+3\sqrt[3]{\left(26+15\sqrt[]{3}\right)\left(26-15\sqrt[]{3}\right)}.x\)

\(\Leftrightarrow x^3=52+3x\)

\(\Leftrightarrow x^3-3x-52=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+13\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left[\left(x+2\right)^2+9\right]=0\)

\(\Leftrightarrow x=4\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)

\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)

\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)

a)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)

\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)

\(\Leftrightarrow\sqrt{1}=1\) (đpcm)

\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}-1=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{3}}-1=0\)

\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)-1=0\)

\(\Leftrightarrow\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2-1=0\)

\(\Leftrightarrow3-2-1=0\) (đpcm)

Yêu cầu đề bài là gì em?

rút gọn những biểu thức sau 

Lời giải:

Gọi biểu thức trên là $A$

Đặt \(\sqrt[3]{15\sqrt{3}-26}=a; \sqrt[3]{15\sqrt{3}+26}=b\). Ta có:

\(a^3-b^3=-52\)

\(ab=-1\)

\(A^3=(a-b)^3=a^3-3ab(a-b)-b^3=-52+3A\)

\(\Leftrightarrow A^3-3A+52=0\)

\(\Leftrightarrow A^2(A+4)-4A(A+4)+13(A+4)=0\)

\(\Leftrightarrow (A+4)(A^2-4A+13)=0\)

Dễ thấy $A^2-4A+13>0$ nên $A+4=0$

$\Leftrightarrow A=-4$

Ta có: \(\hept{\begin{cases}\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)=2+\sqrt{3}\\\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)=2-\sqrt{3}\end{cases}}\)

Sửa đề:

\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

\(=\frac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\)

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

\(=\sqrt{2}\)