a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)

\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)

\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)

hay \(B=2\sqrt{10}\)

d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)

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

hay \(D=\sqrt{2}\)

a: \(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

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

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

\(=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)

a,=3.007298903

b,=2.732050808

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

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

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

\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-\left(2+\sqrt{3}\right)}\)(Vì \(2>\sqrt{3}>0\))

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

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

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

\(=\sqrt{4-3}\)(Vì \(3>0\))

\(=1\)

Ta có: \(B=\left(\frac{1}{\sqrt{\sqrt{5}+2}}+\sqrt{\sqrt{5}+2}\right)\cdot\frac{1}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

\(=\left(\frac{1}{\sqrt{\sqrt{5}+2}}+\frac{\sqrt{5}+2}{\sqrt{\sqrt{5}+2}}\right)\cdot\frac{\sqrt{\sqrt{5}-1}}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)

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

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

\(=\frac{3+\sqrt{5}}{2}\cdot\sqrt{\frac{7-3\sqrt{5}}{5-4}}-\left(\sqrt{2}-1\right)\)(Vì \(\sqrt{2}>1\))

\(=\frac{6+2\sqrt{5}}{4}\cdot\sqrt{7-3\sqrt{5}}-\sqrt{2}+1\)

\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\sqrt{14-6\sqrt{5}}}{4\sqrt{2}}-\sqrt{2}+1\)

\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\left(3-\sqrt{5}\right)}{4\sqrt{2}}-\frac{4\sqrt{2}\cdot\left(\sqrt{2}-1\right)}{4\sqrt{2}}\)

\(=\frac{\left(\sqrt{5}+1\right)^2\cdot\left(6-2\sqrt{5}\right)}{8\sqrt{2}}-\frac{8\sqrt{2}\cdot\left(\sqrt{2}-1\right)}{8\sqrt{2}}\)

\(=\frac{16-16+8\sqrt{2}}{8\sqrt{2}}\)

=1

Vậy: A=B

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

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

\(=4-3=1\)

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

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

\(=12-5=7\)

a) (2 - √3)(2 + √3)

= 2² - (√3)²

= 4 - 3

= 1

b) (2√3 - √5)(2√3 + √5)

= (2√3)² - (√5)²

= 12 - 5

= 7

Bài 2:

Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)

\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)

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

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

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

=1

câu 1. đkxđ: \(x\ge\frac{1}{2}\)
\(A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)

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

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

\(=\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|\)

nếu \(\left|\sqrt{2x-1}-1\right|=\sqrt{2x-1}-1\) với \(\sqrt{2x-1}\ge1\Leftrightarrow x\ge1\)

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

=> A=\(\sqrt{2}\)

nếu \(\left|\sqrt{2x-1}-1\right|=1-\sqrt{2x-1}\) với \(\frac{1}{2}\le x< 1\)

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

=> A= \(\sqrt{4x-2}\)

câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)

\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)

=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)

\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)

\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)

= 4

\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)

\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)

\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)

\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)

\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)

https://hoc24.vn/hoi-dap/question/405366.html

\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

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

= 5 - 3

= 2