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

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

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

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

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

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

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

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

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

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

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

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

\(=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

\(B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)

\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)

\(=\sqrt{2}.\sqrt{4-2}=\sqrt{2}.\sqrt{2}=2\)

\(C=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

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

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

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

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

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

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

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

\(E=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right).\sqrt{3-\sqrt{5}}\)

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

\(=2\sqrt{3-\sqrt{5}}+2\sqrt{3+\sqrt{5}}=\sqrt{2}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)

\(=\sqrt{2}.\left(\sqrt{5}-1+\sqrt{5}+1\right)=2\sqrt{10}\)

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 \(D=\sqrt{2}\)

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}\)\(-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}\)\(+\sqrt{5}.\sqrt{3}\)\(-\sqrt{4.15}\)

\(=2.3+\sqrt{15}-2\sqrt{15}\)

\(=6+\sqrt{15}.\left(1-2\right)\)

\(=6-\sqrt{15}\)

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=2.3+\sqrt{5.3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)

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

\(\left(2+\sqrt{3}\right)\left(\sqrt{7-4\sqrt{3}}\right)=\left(2+\sqrt{3}\right)\sqrt{4-4\sqrt{3}+3}\)

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

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

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

\(=4-2=1\)

mk

2.+ \(\left(2n+1\right)^2=4n^2+4n+1>4n^2+4n\)

\(\Rightarrow2n+1>\sqrt{4n\left(n+1\right)}=2\sqrt{n\left(n+1\right)}\)

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

\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< \frac{1}{2}\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)

\(\Rightarrow A< \frac{1}{2}\)

1. + \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)-n}{\left(n+1\right)\sqrt{n}}=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\left(n+1\right)\sqrt{n}}\)

\(< \frac{\left(\sqrt{n+1}-\sqrt{n}\right)\cdot2\sqrt{n+1}}{\sqrt{n}\left(n+1\right)}=2\cdot\frac{n+1-\sqrt{n\left(n+1\right)}}{\left(n+1\right)\sqrt{n}}=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Do đó : \(A< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\right)\)

\(\Rightarrow A< 2\)

Bài 2 tạm thời chưa nghĩ ra :))