Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{2015-2014}{\sqrt{2015}+\sqrt{2014}}>\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\sqrt{2016}-\sqrt{2015}\)

Ta có:  √2015−√2014=2015−2014√2015+√2014>2016−2015√2016+√2015=√2016−√2015

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a \(\ne\) b ta có:

\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{2015}\)

\(\Rightarrow\sqrt{2016}+\sqrt{2014}< 2.\sqrt{2015}\)

\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

sprt là gì

bằng nhau. vì

= sqrt(2017-2016) =sqrt (1)

=sqrt(2016-2015) =sqrt (2)

từ (1) (2) => 2 cái đó bằng nhau.

đây là cách trình  bày nháp. khi bạn viết ra bài thì ghi  đề ra nha. CHÚC HỌC TỐT!

\(\sqrt{2017}-\sqrt{2016}\) với \(\sqrt{2016}-\sqrt{2015}\)

Ta có : 

\(\sqrt{2017}-\sqrt{2016}=\frac{2017-2016}{\sqrt{2017}+\sqrt{2016}}\) \(=\frac{1}{\sqrt{2017}+\sqrt{2016}}< \frac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2016}-\sqrt{2015}=\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\)

Do đó :
\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)

Chúc bạn học tốt !!!

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

1) Ta có bđt sau : \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m)

Áp dụng : \(\frac{\sqrt{2005}+\sqrt{2007}}{2}< \sqrt{\frac{2005+2007}{2}}\)

\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)

2) Xét : \(A-B=2\sqrt{2014}-\left(\sqrt{2013}+\sqrt{2015}\right)\)

Theo câu 1) , ta dễ dàng c/m được \(2\sqrt{2014}>\sqrt{2013}+\sqrt{2015}\)

Do đó A - B > 0 => A > B

2) Bình phương 2 vế ta có:

 \(A^2=2014-2013=1\)

\(B^2=2015-2014=1\)

=>A=B

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2014}+\sqrt{2015}\)

nên \(\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

bình từng cái @

\(\dfrac{1}{7}\sqrt{51}với\dfrac{1}{9}\sqrt{150}\)

<=> \(\dfrac{\sqrt{51}}{7}với\dfrac{\sqrt{150}}{9}\)

<=> \(9\sqrt{51}với7\sqrt{150}\)

<=> \(\sqrt{4131}với\sqrt{7350}\)

=> \(\sqrt{4131}< \sqrt{7350}\)

=> \(\dfrac{1}{7}\sqrt{51}< \dfrac{1}{9}\sqrt{150}\)