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1.
Đặt biểu thức là $A$
Ta thấy:
$\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$
Tương tự với các phân số còn lại và công theo vế thì:
$A=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+...+(\sqrt{2019}-\sqrt{2018})$
$=\sqrt{2019}-1$
2.
$\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{5.3}+3}+\sqrt{3-2\sqrt{3.1}+1}$
$=\sqrt{(\sqrt{5}-\sqrt{3})^2}+\sqrt{(\sqrt{3}-1)^2}$
$=|\sqrt{5}-\sqrt{3}|+|\sqrt{3}-1|$
$=\sqrt{5}-\sqrt{3}+\sqrt{3}-1=\sqrt{5}-1$
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
Ta có: \(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)^2-k^2\left(k+1\right)}\)
\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k^3+2k^2+k-k^3-k^2}\)
\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k\left(k+1\right)}\)
\(=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Lần lượt thay k=1;2;...;2018 ta được:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{1}-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
...
\(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}=\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
Cộng vế theo vế ta được:
\(C=1-\frac{1}{\sqrt{2019}}=...\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{4}}-...-\frac{1}{\sqrt{2018}-\sqrt{2019}}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{2019}-\sqrt{2018}}{2019-2018}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}\)
\(=\sqrt{2019}-\sqrt{2}\)
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(n+1-n)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
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?