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1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
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:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
\(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}\ge\frac{\left(\sqrt{2019}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}\)
Dấu "=" ko xảy ra nên \(\frac{2019}{\sqrt{2018}}+\frac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)
\(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3\left(\sqrt{5}+1\right)}{5-1}\)
\(=\dfrac{14}{5}\sqrt{5}+\dfrac{3}{4}\sqrt{5}+\dfrac{3}{4}\)
\(=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)
a: \(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3}{4}+\dfrac{3}{4}\sqrt{5}=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)
b: Đặt a=2018
\(B=\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}\)
\(=\sqrt{a^2+\left(a^2+a\right)^2+a^2+2a+1}\)
\(=\sqrt{2a^2+1+2a+a^4+2a^3+a^2}\)
\(=\sqrt{a^4+2a^3+3a^2+2a+1}\)
\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1=2018^2+2018+1\)
a) Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=4036+2\sqrt{\left(2018-1\right).\left(2018+1\right)}\)
\(=4036+2\sqrt{2018^2-1}< 4036+2\sqrt{2018^2}=2018.4=\left(2\sqrt{2018}\right)^2\)
Vậy x < y
Ta có:
\(\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}=\dfrac{2018}{\sqrt{2018}}+\dfrac{1}{\sqrt{2018}}+\dfrac{2019}{\sqrt{2019}}-\dfrac{1}{\sqrt{2019}}=\sqrt{2018}+\sqrt{2019}+\left(\dfrac{1}{\sqrt{2018}}-\dfrac{1}{\sqrt{2019}}\right)\)
Do \(\dfrac{1}{\sqrt{2018}}>\dfrac{1}{\sqrt{2019}}\) nên \(\dfrac{1}{\sqrt{2018}}-\dfrac{1}{\sqrt{2019}}\) dương \(\Rightarrow\dfrac{2019}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2019}}>\sqrt{2018}+\sqrt{2019}\)
20192018−−−−√+20182019−−−−√=20182018−−−−√+12018−−−−√+20192019−−−−√−12019−−−−√=2018−−−−√+2019−−−−√+(12018−−−−√−12019−−−−√)20192018+20182019=20182018+12018+20192019−12019=2018+2019+(12018−12019)
Do 12018−−−−√>12019−−−−√12018>12019 nên 12018−−−−√−12019−−−−√12018−12019 dương ⇒20192018−−−−√+20182019−−−−√>2018−−−−√+2019−−−−√