c) Bình phương hai vế ta được 2015+2017+2\(\sqrt{2015\times2017}\) và 4\(\times\)2016
Ta có 2015 + 2017 + 2\(\sqrt{2015\times2017}\)
= (2016-1) + (2016+1) + 2\(\sqrt{2015\times2017}\)
= 2016 + 2016 + 1 - 1 + 2\(\sqrt{2015\times2017}\)
= 2\(\times\)2016 + 2\(\sqrt{2015\times2017}\) (1)
ta thấy 2015 \(\times\) 2017 =(2016-1) \(\times\) (2016+1)= 20162 - 1
nên (1) \(\Leftrightarrow\)2\(\times\)2016 + 2\(\sqrt{2016^2-1}\)
Ta có 4\(\times\)2016=2\(\times\)2016 + 2\(\times\)2016=2\(\times\)2016 + 2\(\sqrt{2016^2}\)
Vì 20162-1 < 20162 nên 2\(\sqrt{2016^2-1}\) < 2\(\sqrt{2016^2}\)
\(\Leftrightarrow\) 2\(\times\)2016 + 2\(\sqrt{2016^2-1}\) < 2\(\times\)2016 + 2\(\sqrt{2016^2}\)
\(\Leftrightarrow\)2015+2017+2\(\sqrt{2015\times2017}\) < 4\(\times\)2016
Hay \(\sqrt{2015}+\sqrt{2017}\) < \(2\sqrt{2016}\)
a) Bình phương hai vế ta được 5+7+\(2\sqrt{5\times7}\) và 13.
Ta có 5+7+\(2\sqrt{5\times7}\) =12+\(2\sqrt{35}\)
13=12+1=12+\(2\times\frac{1}{2}\) =12+\(2\sqrt{\frac{1}{4}}\)
Vì 35 > \(\frac{1}{4}\) nên \(\sqrt{35}\) > \(\sqrt{\frac{1}{4}}\) \(\Leftrightarrow\)2\(\sqrt{35}\) > \(2\sqrt{\frac{1}{4}}\) \(\Leftrightarrow\)12+2\(\sqrt{35}\) > 12+\(2\sqrt{\frac{1}{4}}\)
Hay\(\sqrt{5}\)+\(\sqrt{7}\) > \(\sqrt{13}\)
b) Bình phương hai vế ta được 162 và 15\(\times\)17
Ta có 15\(\times\)17=(16-1)\(\times\)(16+1)=162-1 < 162
\(\Leftrightarrow\)162 > 15\(\times\)17 Hay \(\sqrt{16}>\sqrt{15}\times\sqrt{17}\)
b, Có \(\sqrt{15}.\sqrt{17}=\sqrt{16-1}.\sqrt{16+1}=\sqrt{16^2-1}< \sqrt{16^2}=16\)
=> \(\sqrt{15}.\sqrt{17}< 16\)
c,Có \(\left(\sqrt{2015}+\sqrt{2017}\right)^2=2015+2017+2\sqrt{2015.2017}=4032+2\sqrt{2015.2017}=4032+2\sqrt{\left(2016-1\right)\left(2016+1\right)}\)
= \(4032+2\sqrt{2016^2-1}\) (1)
\(\left(2\sqrt{2016}\right)^2=4.2016=4032+2.2016=4032+2\sqrt{2016^2}>4032+2\sqrt{2016^2-1}\)(2)
Từ (1),(2) => \(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\)
<=> \(\sqrt{2015}+\sqrt{2016}< 2\sqrt{2016}\)
