a) \(\sqrt{19+\sqrt{136}}\) -\(\sqrt{19-\sqrt{136}}\)
= \(\sqrt{19+2\sqrt{34}}\) - \(\sqrt{19-2\sqrt{34}}\)
= \(\sqrt{\left(\sqrt{17}+\sqrt{2}\right)^2}\) - \(\sqrt{\left(\sqrt{17}-\sqrt{2}\right)^2}\)
= \(\left|\sqrt{17}+\sqrt{2}\right|-\left|\sqrt{17}-\sqrt{2}\right|\)
= \(\sqrt{17}+\sqrt{2}-\sqrt{17}+\sqrt{2}\)
= \(2\sqrt{2}\)

Công thức viết khó đọc quá. Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.

Đặt \(x=\sqrt{5+\sqrt{19+2\sqrt{5}}}-\sqrt{5-\sqrt{19+2\sqrt{5}}}>0\)

\(x^2=10-2\sqrt{\left(5+\sqrt{19+2\sqrt{5}}\right)\left(5-\sqrt{19+2\sqrt{5}}\right)}\)

\(x^2=10-2\sqrt{6-2\sqrt{5}}\)

\(x^2=10-2\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(x^2=10-2\sqrt{5}+2\)

\(x^2=12-2\sqrt{5}\)

\(\Rightarrow x=\sqrt{12-2\sqrt{5}}\)

Sqrt ở đây làm căn bậc hai nhé

\(D=\left|\sqrt{8}-3\right|+\left|\sqrt{19}-4\right|-\left(\sqrt{19}-\sqrt{8}\right)\)

\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-\sqrt{8}\right)\)

\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)

\(D=-2\sqrt{2}+3+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)

\(D=-2\sqrt{2}+3+\sqrt{19}-4-\sqrt{19}+2\sqrt{2}\)

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

\(D=3-4\)

\(D=-1\)

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

\(=16-3+3-16=0\)

Lời giải:

Đặt \(\sqrt{\sqrt{20}+\sqrt{19}}=a\)

Ta thấy:

\(\sqrt{\sqrt{20}+\sqrt{19}}.\sqrt{\sqrt{20}-\sqrt{19}}=\sqrt{(\sqrt{20}-\sqrt{19})(\sqrt{20}+\sqrt{19})}=\sqrt{20-19}=1\)

\(\Rightarrow \sqrt{\sqrt{20}-\sqrt{19}}=\frac{1}{\sqrt{\sqrt{20}+\sqrt{19}}}=\frac{1}{a}\)

PT trở thành:

\(a^x+\frac{1}{a^x}=2\)

\(\Leftrightarrow a^{2x}+1-2a^{x}=0\)

\(\Leftrightarrow (a^x-1)^2=0\Rightarrow a^x=1\)

Mà \(a\neq 1\) nên \(\Rightarrow x=0\)

a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)

Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)