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\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)
\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)
....... Tới đây được chưa bạn?
a) Ta có :\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\) =\(\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Tương tự : \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\) = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
=>\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)+\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)= \(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)=\(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{3-2}\)=10
Ta có: \(\frac{1}{3}\left(\sqrt{6}+\sqrt{5}\right)^2-\frac{1}{4}\sqrt{120}-2\sqrt{\frac{15}{2}}\)
\(=\frac{1}{3}\left(11+2\sqrt{30}\right)-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}+\frac{2}{3}\sqrt{30}-\frac{\sqrt{30}}{2}-\sqrt{30}\)
\(=\frac{11}{3}-\frac{5}{6}\sqrt{30}\)
\(=\frac{22-5\sqrt{30}}{6}\)
Ta có: \(\left(\frac{1}{2}\sqrt{\frac{2}{3}}-\frac{3}{4}\sqrt{54}+\frac{1}{3}\sqrt{\frac{8}{3}}\right)\div\sqrt{\frac{81}{6}}\)
\(=\left(\frac{\sqrt{6}}{6}-\frac{9\sqrt{6}}{4}+\frac{2\sqrt{6}}{9}\right)\div\frac{3\sqrt{6}}{2}\)
\(=-\frac{67\sqrt{6}}{36}\cdot\frac{2}{3\sqrt{6}}\)
\(=-\frac{67}{54}\)
\(=\frac{2-1}{\sqrt{2}+1}+\frac{3-2}{\sqrt{3}+\sqrt{2}}+\frac{4-3}{\sqrt{4}+\sqrt{3}}+...+\frac{100-99}{\sqrt{100}+\sqrt{99}}.\)
\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2}+1}+\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}+\frac{\left(\sqrt{4}+\sqrt{3}\right)\left(\sqrt{4}-\sqrt{3}\right)}{\sqrt{4}+\sqrt{3}}+...\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=10-1=9.\)
a, \(\frac{\sqrt{2}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\)-\(\frac{3\left(1+\sqrt{3}\right)}{1+\sqrt{3}}\)
=\(\sqrt{2}-3\)
b,X=\(\sqrt{2019}+\sqrt{2018}\)
(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2019}+\sqrt{2018}\))
Y=\(\sqrt{2018}+\sqrt{2017}\)
(Khử mẫu,nhân tử&mẫu vs\(\sqrt{2018}+\sqrt{2017}\))
So sánh:X & Y<=>X-\(\sqrt{2018}\)&Y-\(\sqrt{2018}\)(Trừ hai vế cho \(\sqrt{2018}\)) <=>\(\sqrt{2019}\)&\(\sqrt{2017}\)
Có:2019>2017
=>\(\sqrt{2019}>\sqrt{2017}\)
=>X>Y
Câu b, mk ko bt có lm đúng ko?
Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{168}+\sqrt{169}}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{169}-\sqrt{168}}{\left(\sqrt{168}+\sqrt{169}\right)\left(\sqrt{169}-\sqrt{168}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{169}-\sqrt{168}}{169-168}\)
\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{169}-\sqrt{168}\)
\(A=\sqrt{169}-\sqrt{1}\)
\(A=13-1=12\)( đpcm )