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

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

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

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

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

ok  mk giải dk tối qua rồi , dù s cx thanks

Thế muốn giải thích thì liệt kê đau đầu =(

\(\frac{3}{\sqrt{7}-5}-\frac{3}{\sqrt{7+5}}=\frac{-10}{9}\inℚ\)

\(\frac{\sqrt{7}+5}{\sqrt{7}-5}+\frac{\sqrt{7}-5}{\sqrt{7}+5}=12\inℚ\)

Đây là TH là số hữu tỉ còn lại.....

\(\frac{4}{2-\sqrt{3}}-\frac{4}{2+\sqrt{3}}=8\sqrt{3}\notinℚ\)

\(\frac{\sqrt{3}}{\sqrt{7}-2}-2\sqrt{7}=2-\sqrt{7}\notinℚ\)

E = \(6x+\sqrt{9x^2-12x+4}\)

E = \(6x+\sqrt{\left(3x-2\right)^2}\)

E = \(6x+\left|3x-2\right|\)

E = \(6x+3x-2\)

E = \(9x-2\)

F = \(5x-\sqrt{x^2+4x+4}\)

F = \(5x-\sqrt{\left(x+2\right)^2}\)

F = \(5x-\left|x+2\right|\)

F = \(5x-x+2\)

F = \(4x+2\)

A. -0,8 ×0,125=-0,1

b. 2^3+3^2=8+9=17

c.=1

d.=-2

\(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2-\sqrt{3}\right)^2}}}\)

\(=\sqrt{5\sqrt{3}+5\sqrt{48-20+10\sqrt{3}}}\)

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

\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)

= 5

\(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)

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

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

\(=3\)