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

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

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

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

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:

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

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

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

\(=\sqrt{2}\)

a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

\(=5\sqrt{6}\)

b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

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

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

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

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

\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

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

\(a^3=30\sqrt{3}+3a.\sqrt[3]{15^2.3-26^2}=30\sqrt{3}-3a\)

\(\Leftrightarrow a^3+3a-30\sqrt{3}=0\)

\(\Leftrightarrow\left(a-2\sqrt{3}\right)\left(a^2+2\sqrt{3}a+15\right)=0\)

\(\Rightarrow a=2\sqrt{3}\)

Ta có: \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\frac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

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

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

\(=\frac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)(Vì \(3\sqrt{3}>5>0\))

\(=\frac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

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

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

\(=\frac{2}{\sqrt{2}}=\sqrt{2}\)