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câu b trc nha
B = \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{6}+\sqrt{8}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{4+\sqrt{2}-\sqrt{3}-\sqrt{2}.\sqrt{3}+2\sqrt{2}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{2+2+\sqrt{2}+2\sqrt{2}-\sqrt{3}-\sqrt{6}}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)+2\left(\sqrt{2}+1\right)-\sqrt{3}\left(\sqrt{2}+1\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\dfrac{\left(\sqrt{2}+1\right)\left(2+\sqrt{2}-\sqrt{3}\right)}{2+\sqrt{2}-\sqrt{3}}\)
= \(\sqrt{2}\) + 1
A = \(\dfrac{21}{2}\) . (\(\sqrt{4+2\sqrt{3}}\) + \(\sqrt{6-2\sqrt{5}}\) )2 - 15\(\sqrt{15}\)
- 3(\(\sqrt{4-2\sqrt{3}}\) +\(\sqrt{6+2\sqrt{5}}\) )2
= \(\dfrac{21}{2}\).(\(\sqrt{\left(\sqrt{3}+1\right)^2}\) + \(\sqrt{\left(\sqrt{5}-1\right)^2}\))2-15\(\sqrt{15}\)
-3(\(\sqrt{\left(\sqrt{3}-1\right)^2}\) + \(\sqrt{\left(\sqrt{5}+1\right)^2}\))2
= \(\dfrac{21}{2}\).(\(\sqrt{3}\) +1+ \(\sqrt{5}\) - 1)2 -3.(\(\sqrt{3}\) - 1 + \(\sqrt{5}\) +1)2
- 15\(\sqrt{15}\)
= \(\dfrac{21}{2}\).(8+2\(\sqrt{15}\) ) - 3(8 + 2\(\sqrt{15}\) ) -15\(\sqrt{15}\)
= \(\dfrac{15}{2}\) .2.(4+\(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.( 4 + \(\sqrt{15}\) ) - 15\(\sqrt{15}\)
= 15.(4+\(\sqrt{15}\) -\(\sqrt{15}\)) =15.4 = 60
Vậy A = 60.
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
Làm luôn nhé
\(2B=21.2\left[\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)\right]^2-2.15\sqrt{15}\)
\(2B=21\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-6\left(\sqrt{3}-1+\sqrt{5}-1\right)^2-30\sqrt{15}\)
\(2B=21\left(\sqrt{3}+\sqrt{5}\right)^2-6\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(\sqrt{3}+\sqrt{5}\right)^2-30\sqrt{15}\)
\(2B=15\left(8+2\sqrt{15}\right)-30\sqrt{15}\)
\(2B=120+30\sqrt{15}-30\sqrt{5}\)
\(2B=120\)
\(B=60\)