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\(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2019+2\sqrt{2017.2019}\)
\(=2.2018+2\sqrt{2018^2-1}< 2.2018+2.2018=4.2018\)
Ta có: \(\left(\sqrt{2017}+\sqrt{2019}\right)^2< 4.2018\)
\(\Rightarrow\sqrt{2017}+\sqrt{2018}< 2.\sqrt{2018}\)
Tham khảo nhé~
Bài 1: Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{1}{\sqrt{2020}+\sqrt{2019}};\)\(\sqrt{2018}-\sqrt{2017}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Dễ thấy \(\sqrt{2020}+\sqrt{2019}>\sqrt{2018}+\sqrt{2017}\)nên \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)
Suy ra\(\sqrt{2020}-\sqrt{2019}< \sqrt{2018}-\sqrt{2017}\)
Bài 2: Xét biểu thức \(\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}=\sqrt{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)(Vì \(a^2+a+1>0\forall a\inℝ\))
Áp dụng công thức tổng quát trên, ta được: \(\sqrt{2019^2+2019^2.2020^2+2020^2}=2019^2+2019+1\)(là số tự nhiên) (đpcm)
Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)
\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)
Vậy A<B
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
cả hai bài đều giải bằng cách bình phương cả hai vế rồi so sánh
So sánh từng vế:
\(\sqrt{15}+1=4,872983346\)
\(\sqrt{24}=4,898979486\)
Vậy: \(\sqrt{15}+1< \sqrt{24}\)
\(\sqrt{2002}+\sqrt{2004}=89,50977321\)
\(2\sqrt{2005}=89,5545271\)
Vậy \(\sqrt{2002}+\sqrt{2004}< 2\sqrt{2005}\)
P/s: Ko chắc
\(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015\cdot2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016\cdot2017}\)
\(2015\cdot2018=2015\cdot2017+2015=2017\cdot\left(2015+1\right)-2017+2015\)
\(=2017\cdot2016-2\)
\(\Rightarrow2015\cdot2018< 2016\cdot2017\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\)
a, x=\(\frac{1\left(\sqrt{2019}+\sqrt{2018}\right)}{2019-2018}\) và y=\(\frac{1\left(\sqrt{2018}+\sqrt{2017}\right)}{2018-2017}\) (Trục căn thức ở mẫu)
\(\Leftrightarrow\) x=\(\sqrt{2019}+\sqrt{2018}\) và y=\(\sqrt{2018}+\sqrt{2017}\)
b, Ta có : x - y = (\(\sqrt{2019}+\sqrt{2018}\) ) - ( \(\sqrt{2018}+\sqrt{2017}\) )
= \(\sqrt{2019}-\sqrt{2017}\) > 0
\(\Rightarrow\) x - y > 0 \(\Leftrightarrow\) x > y
\(\left(\sqrt{2017}+\sqrt{2019}\right)^2=2017+2\sqrt{4072323}+2019=4036+2\sqrt{4072323}=4036+\sqrt{16289292}\)
\(\left(2\sqrt{2018}\right)^2=4.2018=4036+4036=4036+\sqrt{16289296}=4036+\sqrt{16289296}\)
Vì \(16289292< 16289296=>\sqrt{16289292}< \sqrt{16289296}\)
\(=>\left(\sqrt{2017}+\sqrt{2019}\right)^2< 2\sqrt{2018}\left(đpcm\right)\)
hàng cuối ghi nhầm r
cảm ơn bn nha