So Sánh :
\(\frac{2015}{\sqrt{2016}}\) và \(\frac{2016}{\sqrt{2015}}\)
So sánh:
A=\(\frac{2015}{\sqrt{2016}}+\frac{2016}{\sqrt{2015}}\) và B=\(\sqrt{2015}+\sqrt{2016}\)
Có: \(\sqrt{2015}< \sqrt{2016}\)
=>\(\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2016}}\)
=>\(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>0\)
=>\(\sqrt{2015}+\sqrt{2016}+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
=>\(\left(\sqrt{2015}+\frac{1}{\sqrt{2015}}\right)+\left(\sqrt{2016}-\frac{1}{\sqrt{2016}}\right)>\sqrt{2015}+\sqrt{2016}\)
=>\(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)
So sánh:
A=\(\frac{2015}{\sqrt{2016}}+\frac{2016}{\sqrt{2015}}\) và B=\(\sqrt{2015}+\sqrt{2016}\)
So sánh: \(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}\text{ và }\sqrt{2015}+\sqrt{2016}\)(không dùng máy tính cầm tay)
So sánh hai bbiểu thức sau \(A=\frac{2014}{\sqrt{2015}}+1vàB=\frac{2015+\sqrt{2016}}{\sqrt{2016}}\)
Ta đặt \(x=2015\), khi đó \(2014=x-1,2016=x+1.\) Ta có như sau
\(2014^2\cdot2016=\left(x-1\right)^2\left(x+1\right)=\left(x^2-1\right)\left(x-1\right)\)\(
so sánh \(\sqrt{2015}-\sqrt{2014}\) và \(\sqrt{2016}-\sqrt{2015}\)
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
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
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
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
Không dùng máy tính, hãy so sánh \(\sqrt{2017}-\sqrt{2016}\) và \(\sqrt{2016}-\sqrt{2015}\)
\(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2017}+\sqrt{2016}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
2017>2015
=>căn 2017>căn 2015
=>\(\sqrt{2017}+\sqrt{2016}>\sqrt{2016}+\sqrt{2015}\)
=>\(\dfrac{1}{\sqrt{2017}+\sqrt{2016}}< \dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
=>\(\sqrt{2017}-\sqrt{2016}< \sqrt{2016}-\sqrt{2015}\)
so sánh (ko dùng bảng số hay máy tính cầm tay):
a) \(\frac{1}{7}\sqrt{51}với\frac{1}{9}\sqrt{150}\)
b) \(\sqrt{2017}-\sqrt{2016}với\sqrt{2016}-\sqrt{2015}\)
So sánh ; \(\sqrt{2016}-\sqrt{2015}và\sqrt{2015}-\sqrt{2014}\)
Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a \(\ne\) b ta có:
\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)
\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)
\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{2015}\)
\(\Rightarrow\sqrt{2016}+\sqrt{2014}< 2.\sqrt{2015}\)
\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)