1: \(\left(\sqrt{10}-\sqrt{14}\right)\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{5}-\sqrt{7}\right)\cdot\sqrt{12+2\sqrt{35}}\)

\(=\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)\)

=5-7=-2

2: Sửa đề: \(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)

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

\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)

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

Sửa đề: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

Ta có: \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

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

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

=1

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

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

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

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

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

Chắc cái căn ở số 2 thứ 4 phải bao trùm luôn chứ nhỉ? nếu ko nó sẽ ko tồn tại quy luật dẫn tới ... ở đằng sau

Đặt \(x=\sqrt{2+\sqrt{2+\sqrt{2+...}}}>0\)

\(\Rightarrow x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}\)

\(\Rightarrow x^2=2+x\)

\(\Rightarrow x^2-x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1< 0\left(ktm\right)\\x=2\end{matrix}\right.\)

Vậy \(x=2\)

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

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

=1

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

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

e có phát hiện mới:v cj chung lớp vs cj kia đúng hemm:v

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

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

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

MATH ERROR

1.994592112

Math error

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