Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

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

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

\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

Suy ra VT = VP = 1

Đáp án là : VT = VP = 1

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ai hâm mộ dĩnh tỷ nhớ kết bạn với mình nhé

a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)

Ta có :

A= \(\dfrac{2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

Đặt B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

Ta có B=\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12+2\cdot\sqrt{12}+1}}}\)

\(2\cdot\sqrt{3+\sqrt{5-\sqrt{12}-1}}\\ =2\sqrt{3+\sqrt{4-\sqrt{12}}}\\ =2\cdot\sqrt{3+\sqrt{3-2\cdot\sqrt{3}+1}}\\ =2\cdot\sqrt{3+\sqrt{3}-1}\\ =2\cdot\sqrt{2+\sqrt{3}}\)

Thay B vào A, ta cũng có:

A=\(\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\ =\dfrac{2\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2\cdot\left(\sqrt{3}+1\right)}}\\ =\dfrac{\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\\ =\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}=\dfrac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}=1\)

Vậy A thuộc Z

Biến đổi vế trái ta có:

\(\sqrt{3+\sqrt{5}-\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{5}-\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{3+\sqrt{5}-2\sqrt{3}-1}=\sqrt{2+\sqrt{5}-2\sqrt{3}}\)

Đề sai 

Trả lời:

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)

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

\(A=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

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

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

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

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

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

\(A=1\)

\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}-1\)

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

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

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

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