1) \(=2\sqrt{5}-3+5-2\sqrt{5}=2\)

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

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

\(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(=\sqrt{\frac{6+2\sqrt{5}}{2}}+\sqrt{\frac{14-6\sqrt{5}}{2}}-\sqrt{2}\)

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

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

\(=2\sqrt{2}-\sqrt{2}\)

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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

\(A=\left(8+2\cdot3-7\cdot\dfrac{13}{10}+3\cdot\dfrac{5}{4}\right):\left(\dfrac{5\sqrt{6}}{3}\right)^2\\ A=\left(14-\dfrac{91}{10}+\dfrac{15}{4}\right):\dfrac{50}{3}\\ A=\dfrac{173}{20}\cdot\dfrac{3}{50}=\dfrac{519}{1000}\)

a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)

=3

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

Chiều cao là

     288 x 2 : 32 = 18 ( m )

Diện tích mới là

      288 + 72 = 360 ( m2 )

Cạnh đáy mới là 360 x 2 : 18 = 40 ( m2 )

Tăng cạnh đáy thêm 40 - 32 = 8 ( m )

1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)

\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)

\(=6-8=-2\)

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

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

\(=9-5=4\)

3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

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

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

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

4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)

=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn

5) Đặt \(A=\sqrt{4\sqrt{6}+11}-\sqrt{11-4\sqrt{6}}\)

\(\Rightarrow A^2=\left(\sqrt{11+4\sqrt{6}}-\sqrt{11-4\sqrt{6}}\right)^2\)

\(=11+4\sqrt{6}-2\sqrt{\left(11+4\sqrt{6}\right)\left(11-4\sqrt{6}\right)}+11-4\sqrt{6}\)

\(=22-2\sqrt{121-96}\)

\(=22-2\sqrt{5}\)

=> \(A=\sqrt{22-2\sqrt{5}}\)

6) Đặt \(B=\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\)

\(\Leftrightarrow B^2=\left(\sqrt{10+2\sqrt{11}}-\sqrt{10-2\sqrt{11}}\right)^2\)

\(=10+2\sqrt{11}-2\sqrt{\left(10+2\sqrt{11}\right)\left(10-2\sqrt{11}\right)}+10-2\sqrt{11}\)

\(=20-2\sqrt{100-44}\)

\(=20-4\sqrt{14}\)

=> \(B=\sqrt{20-4\sqrt{14}}\)