Câu 1: 

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

Câu 2: 

\(\Leftrightarrow\left|2x-3\right|=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=2\sqrt{3}\\2x-3=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{3}+3}{2}\\x=\dfrac{-2\sqrt{3}+3}{2}\end{matrix}\right.\)

Đặt \(x=\sqrt{\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}}>0\)

\(x^2=\dfrac{5+2\sqrt{6}}{5-\sqrt{6}}+\dfrac{5-2\sqrt{6}}{5+\sqrt{6}}+2\sqrt{\dfrac{25-24}{25-6}}=\dfrac{74}{19}+\dfrac{2\sqrt{19}}{19}\)

\(\Rightarrow x^2=\dfrac{74+2\sqrt{19}}{19}\Rightarrow x=\sqrt{\dfrac{74+2\sqrt{19}}{19}}\)

Ko thể rút gọn thêm nữa (có thể trục căn thức ở mẫu)

Sửa đề: \(C=21\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2-6\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2-15\sqrt{15}\)

Đặt \(A=\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4+2\sqrt3}+\sqrt{6-2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3+1\right)^2}+\sqrt{\left(\sqrt5-1\right)^2}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt3+1+\sqrt5-1\right)=\frac{1}{\sqrt2}\left(\sqrt3+\sqrt5\right)\)

Đặt \(B=\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}+\sqrt{6+2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt3+\sqrt5\right)\)

Ta có: \(C=21\left(\sqrt{2+\sqrt3}+\sqrt{3-\sqrt5}\right)^2-6\left(\sqrt{2-\sqrt3}+\sqrt{3+\sqrt5}\right)^2-15\sqrt{15}\)

\(=21\left\lbrack\frac{1}{\sqrt2}\left(\sqrt3+\sqrt5\right)\right\rbrack^2-6\cdot\left\lbrack\frac{1}{\sqrt2}\left(\sqrt3+\sqrt5\right)\right\rbrack^2-15\sqrt{15}\)

\(=15\cdot\frac12\left(\sqrt3+\sqrt5\right)^2-15\sqrt{15}=7,5\left(8+2\sqrt{15}\right)-15\sqrt{15}\)

\(=60+15\sqrt{15}-15\sqrt{15}=60\)

a: A=2-căn 3+căn 3=2

b: B=căn 5+1-căn 5+1=2

a: \(E=1+1=2\)

b: \(=6+3\sqrt{5}+\sqrt{6}-\sqrt{2}+\sqrt{6}-\sqrt{5}\)

\(=6+2\sqrt{6}-\sqrt{2}+2\sqrt{5}\)

d: \(=2+\sqrt{3}+2-\sqrt{3}=4\)

= \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

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

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

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

= \(\sqrt{3}+\sqrt{5}+3\)

Bạn Khanh đúng r chỉ sai chỗ\(\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\) mới đúng

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)